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ELEMENTARY   ALGEBRA 


EMBRACING 


THE  FIRST  PRINCIPLES  OF  THE  SCIENCE 


BY 

CHARLES   DAVIES,   LL.D. 

AUTHOR  OF  A  FULL   COURSE   OF  MATHEMATICS 


Edited  by  J.  H.  VAN  AMRINGE,  Ph.D.,  Professor  of  Mathematics, 
Columbia  College 

^       OF  THE     ^         \ 


UNIVERSITY 


OF 


NEW  YORK  • :  •  CINCINNATI  • :  •  CHICAGO 

AMERICAN    BOOK    COMPANY 


Copyright,  1859,  by  Charles  Davies. 
Copyright,  1887,  by  Mary  Ann  Davie8. 
Copyright,  1891,  by  American  Book  Company. 


Typography  by  J.  8.  Cushing  &  Co.,  Boston. 


PREFACE. 


In  the  admirable  series  of  mathematical  text-books,  still 
unequaled  in  many  essential  particulars,  which  the  late 
Professor  Charles  Davies  issued,  the  "  Elementary  Algebra" 
has  an  important  place.  The  intent  and  scope  of  the  work  is 
set  forth  in  the  preface  to  the  latest  edition  that  the  accom- 
plished author  prepared  :  — 

"  Algebra  naturally  follows  arithmetic  in  a  course  of  scien- 
tific studies.  The  language  of  figures,  and  the  elementary 
combinations  of  numbers,  are  acquired  at  an  early  age. 
When  the  pupil  passes  to  a  new  system,  conducted  by 
letters  and  signs,  the  change  seems  abrupt ;  and  he  often 
experiences  much  difficulty  *  before  perceiving  that  algebra 
is  but  arithmetic  written  in  a  different  language. 

"It  is  the  design  of  this  work  to  supply  a  connecting  link 
between  arithmetic  and  algebra  ;  to  indicate  the  unity  of  the 
methods  ;  and  to  conduct  the  pupil  from  the  arithmetical 
processes  to  the  more  abstract  methods  of  analysis,  by  easy 
and  simple  gradations.  The  work  is  also  introductory  to 
the  '  University  Algebra,'  and  to  the  *  Algebra '  of  M.  Bour- 
don, which  latter  is  justly  considered,  both  in  this  country 
and  in  Europe,  as  the  best  text-book  on  the  subject  which  has 
yet  appeared. 


18363 


^j 


4  PREFACE. 

"  In  the  'Introduction,'  or  'Mental  Exercises,'  the  language 
of  figures  and  letters  are  both  employed.  Each  lesson  is 
so  arranged  as  to  introduce  a  single  principle  not  known 
before  ;  and  the  whole  is  so  combined  as  to  prepare  the  pupil, 
by  a  thorough  system  of  mental  training,  for  those  processes 
of  reasoning  which  are  peculiar  to  the   algebraic  analysis." 

The  definitions  are  precise  ;  the  fundamental  principles  and 
operations  are  lucidly  explained  ;  and  the  whole  subject 
within  the  range  of  the  treatise  is  simply  and  logically 
developed,  and  well  illustrated  by  appropriate  examples. 

The  present  edition  is  the  result  of  a  careful  reexamina- 
tion of  the  work,  and  in  it  are  incorporated  such  emendations 
as  the  progress  of  educational  science  and  practical  experience 
in  teaching  have  suggested. 

July,  1891. 


CONTENTS. 

CHAPTER  PAGE 

INTRODUCTION 9 

Mental  Exercises 9 

I.  Definitions 33 

Exercises  in  writing  Algebraic  Expressions 38 

Exercises  in  Interpretation  of  Algebraic  Language     .     .  40 

II.   Fundamental  Operations 41 

Addition 41 

Subtraction 47 

Multiplication 52 

Division 58 

III.  Factoring,  Greatest  Common  Divisor,  and  Least  Com- 

mon Multiple 70 

Useful  Formulas 70 

Factoring 73 

Greatest  Common  Divisor 77 

Least  Common  Multiple 83 

IV.  Fractions 87 

Transformation  of  Fractions 89 

Addition  of  Fractions 95 

Subtraction  of  Fractions .  96 

Multiplication  of  Fractions 98 

Division  of  Fractions 100 

5 


6  CONTENTS. 

CHAPTER  PAGE 

V.  Equations  of  the  First  Degree,  and  Inequalities  .    .  102 

Equations  of  the  First  Degree 102 

Transformation  of  Equations 103 

Solution  of  Equations 106 

Problems  involving  Equations  of  the  First  Degree  .     .  110 

Equations  containing  Two  Unknown  Quantities     .     .  123 

Elimination 124 

Equations  containing  Three  or  more  Unknown  Quan- 
tities    140 

Inequalities 149 

VI.   Powers 153 

Powers  of  Monomials 154 

Powers  of  Fractions 156 

Powers  of  Binomials 158 

VII.   Extraction  of  Roots,  and  Radicals 165 

Extraction  of  Roots 165 

Square  Root 165 

Cube  Root 174 

Any  Root  .     .  " 179 

Radicals 182 

Transformation  of  Radicals 182 

Addition  of  Radicals 190 

Subtraction  of  Radicals 192 

Multiplication  of  Radicals 194 

Division  of  Radicals 196 

Square  Root  of  Polynomials 198 

Cube  Root  of  Polynomials 202 


CONTENTS.  7 

CHAPTER  PAGE 

VIII.   Equations  of  the  Second  Degree 206 

Equations  containing  One  Unknown  Quantity  ....  206 

Incomplete  Equations 208 

Equations  containing  Two  Unknown  Quantities    .     .     .  212 

Complete  Equations 215 

Properties  of  Equations  of  the  Second  Degree     ....  229 

Formation  of  Equations  of  the  Second  Degree    ....  231 

Trinomial  Equations  of  the  Second  Degree 233 

Numerical  Values  of  the  Boots 237 

Equations  containing  more  than  One  Unknown  Quantity,  244 

IX.   Proportions  and  Progressions 258 

Arithmetical  Proportion  and  Progression 258 

Geometrical  Proportion  and  Progression 268 

X.   Logarithms 289 


SUGGESTIONS  TO  TEACHERS. 


1.  The  introduction  is  designed  as  a  mental  exercise.  If 
thorouglily  taught,  it  will  train  and  prepare  the  mind  of  the 
pupil  for  those  higher  processes  of  reasoning  which  it  is  the 
peculiar  province  of  the  algebraic  analysis  to  develop. 

2.  The  statement  of  each  question  should  be  made,  and 
every  step  in  the  solution  gone  through  with,  without  the  aid 
of  a  slate  or  blackboard,  though  perhaps  in  the  beginning 
some  aid  may  be  necessary  to  those  unaccustomed  to  such 
exercises. 

3.  Great  care  must  be  taken  to  have  every  principle  on 
which  the  statement  depends  carefully  analyzed ;  and  equal 
care  is  necessary  to  have  every  step  in  the  solution  distinctly 
explained. 

4.  The  reasoning  process  is  the  logical  connection  of  dis- 
tinct apprehensions,  and  the  deduction  of  the  consequences 
which  follow  from  such  a  connection :  hence  the  basis  of  all 
reasoning  must  lie  in  distinct  elementary  ideas. 

5.  Therefore,  to  teach  one  thing  at  a  time,  to  teach  that 
thing  well,  to  explain  its  connections  with  other  things  and 
the  consequences  which  follow  from  such  connections,  would 
seem  to  embrace  the  whole  art  of  instruction. 


ELEMEISTTARY  ALGEBRA. 

INTRODUCTION. 
MENTAL    EXERCISES. 

Lesson  I. 

1.    John  and  Charles  have  the  same  number  of  apples. 
Both  together  have  12.     How  many  has  each  ? 

Let  X  denote  the  number  which  John  has.  Then,  since  they  have  an 
equal  number,  x  will  also  denote  the  number  which  Charles  has ;  and 
twice  X,  or  2a?,  will  denote  the  number  which  both  have,  which  is  12. 
If  twice  X  is  equal  to  12,  x  will  be  equal  to  12  divided  by  2,  which  is  6 : 
therefore  each  has  6  apples.     Hence  we  write,  — 

Let  X  =  number  of  apples  John  has. 

Then  a;  +  a;  =  2ic  =  12. 

Hence  x  =  —  =    6. 

2 

Note.  —  When  x  is  written  with  the  sign  +  before  it,  it  is  read  plus 
X;  and  the  line  above  is  read,  x  plus  x  equals  12. 

"When  X  is  written  by  itself,  it  is  read  one  x,  and  is  the  same  as 
la:. 

a;  or  1  a;  means  once  x,  or  one     x, 

2x      "       twice  a;,  or  two    x, 

Sx      "      three  times  x,  or  three  a?, 
4  a;      "      four  times    x,  or  four   x,  etc. 

9 


10  ELEMENTARY   ALGEBRA. 

2.  What  IB  x  +  x  equal  to? 

3.  What  is  X  -{-2x  equal  to? 

4.  What  is  x  +  2x  +  X  equal  to ? 

5.  What  is  x-\-bx-\-x  equal  to? 

6.  What  is  ^  + 2a; +  3:r  equal  to? 

7.  James  and  John  together  have  24  peaches,  and  one  * 
has  as  many  as  the  other.     How  many  has  each  ? 

Let  X  denote  the  number  which  James  has.  Then,  since  they  have 
an  equal  number,  x  will  also  denote  the  number  which  John  has ;  and 
twice  X  will  denote  the  number  which  both  have,  which  is  24.  If  twice 
X  is  equal  to  24,  x  will  be  equal  to  24  divided  by  2,  which  is  12 :  there- 
fore each  has  12  peaches.     Hence  we  write, — 

Let  X  =  number  of  peaches  James  has 

Then  a;  +  a;  =  2a;  =  24. 

Hence  a  =  —  =  12. 

2 

A  verification  is  the  operation  of  proving  that  the  num- 
ber found  will  satisfy  the  conditions  of  the  question.     Thus, 

James's  apples.       John's  apples. 

12       +       12     =24. 

Note.  —  Let  the  following  questions  be  analyzed,  written,  and  veri- 
fied in  exactly  the  same  manner  as  the  above. 

8.  William  and  John  together  have  36  pears,  and  one  has 
as  many  as  the  other.     How  many  has  each  ? 

9.  What  number  added  to  itself  will  make  20? 

10.  James  and  John  are  of  the  same  age,  and  the  sum  of 
their  ages  is  32.     What  is  the  age  of  each  ? 

11.  Lucy  and  Ann  are  twins,  and  the  sum  of  their  ages 
is  16.     What  is  the  age  of  each? 

12.  What  number  is  that  which  added  to  itself  will 
make  30  ? 


INTRODUCTION.  11 

13.  What  number  is  that  which  added  to  itself  will 
make  50  ? 

14.  Each  of  two  boys  received  an  equal  sum  of  money  at 
Christmas,  and  together  they  received  60  cents.  How  much 
did  each  receive  ? 

15.  What  number  added  to  itself  will  make  100? 

16.  John  has  as  many  pears  as  William.  Together  they 
have  72.     How  many  has  each  ? 

17.  What  number  added  to  itself  will  give  a  sum  equal 
to  46? 

18.  Lucy  and  Ann  have  each  a  rosebush  with  the  same 
number  of  buds  on  each.  The  buds  on  both  number  46. 
How  many  on  each  ? 

Lesson  II. 

1.  John  and  Charles  together  have  12  apples,  and  Charles 
has  twice  as  many  as  John.     How  many  has  each  ? 

Let  X  denote  the  number  of  apples  which  John  has.  Then,  since 
Charles  has  twice  as  many,  2x  will  denote  his  share;  and  x  +  2x,  or 
3ic,  will  denote  the  number  which  they  both  have,  which  is  12.  If  3  a: 
is  equal  to  12,  x  will  be  equal  to  12  divided  by  3,  which  is  4 :  therefore 
John  has  4  apples  ;  and  Charles,  having  twice  as  many,  has  8.  Hence 
we  write,  — 

Let  X  =  number  of  apples  John  has. 

Then  2x  =  number  of  apples  Charles  has, 

and  (K  -H  2.T  =  3ic  =  12,  the  number  both  have. 

12 
Then  x  =  ~  =    4,  the  number  John  has, 

3 

and  20;  =  2  X  4  =  8,  the  number  Charles  has. 

Verification.     4  +  8  =  12,  the  number  both  have. 

2.  William  and  John  together  have  48  quills,  and  William 
has  twice  as  many  as  John.     How  many  has  each? 


12  ELEMENTARY    ALGEBRA. 

3.  What  number  is  that  which  added  to  twice  itself  will 
give  a  number  equal  to  60  ? 

4.  Charles's  marbles  added  to  John's  make  three  times 
as  many  as  Charles  has.  Together  they  have  51.  How 
many  has  each  ? 

Note.  —  Since  Charles's  marbles  added  to  John's  make  three  times  as 
many  as  Charles  has,  Charles  must  have  one  third,  and  John  two  thirds, 
of  the  whole. 

Let  re  denote  the  number  which  Charles  has.  Then  2x  will  denote 
the  number  which  John  has  ;  and  x  -\-  2x,  or  3  a;,  will  denote  what  they 
both  have,  which  is  51.  Then,  if  3a;  is  equal  to  51,  x  will  be  equal  to 
51  divided  by  3,  which  is  17 :  therefore  Charles  has  17  marbles ;  and 
John,  having  twice  as  many,  has  34.     Hence  we  write, — 

Let  X  =  number  of  Charles's  marbles. 

Then  2x  =  number  of  John's  marbles, 

and  3  a;  =  51,  the  number  of  both. 

51 
Then  a;  =  —  =  17,  Charles's  marbles  ; 

3 

and  2  a;  =  2  X  17  =  34,  John's  marbles. 

5.  What  number  added  to  twice  itself. will  make  75? 

6.  What  number  added  to  twice  itself  will  make  57  ? 

7.  What  number  added  to  twice  itself  will  make  39  ? 

8.  What  number  added  to  twice  itself  will  give  90? 

9.  John  walks  a  certain  distance  on  Tuesday,  twice  as  far 
on  Wednesday,  and  in  the  two  days  he  walks  27  miles.  How 
far  did  he  walk  each  day  ? 

10.  Jane's  bush  has  twice  as  many  roses  as  Nancy's,  and 
on  both  bushes  there  are  36.     How  many  on  each  ? 

11.  Samuel  and  James  bought  a  ball  for  48  cents.  Samuel 
paid  twice  as  much  as  James.     What  did  each  pay  ? 


INTRODUCTION.  13 

12.  Divide  48  into  two  such  parts  that  one  shall  be  double 
the  other. 

13.  Divide  66  into  two  such  parts  that  one  shall  be  double 
the  other. 

14.  The  sum  of  three  equal  numbers  is  12.  What  are  the 
numbers? 

Let  X  denote  one  of  the  numbers.  Then,  since  the  numbers  are  equal, 
X  will  also  denote  each  of  the  others;  and  x  plus  x  plus  x,  or  3a;,  will 
denote  their  sum,  which  is  12.  Then,  if  3  a;  is  equal  to  12,  x  will  be 
equal  to  12  divided  by  3,  which  is  4 :  therefore  the  numbers  are  4,  4, 
and  4.     Hence  we  write,  — 


Let 

X  = 

=  one 

:    of  th( 

3  equal 

numbers. 

Then 

a;  +  a;  +  a;  = 

3x- 

=  12, 

and 

X  = 

.12. 
3  ■ 

=    4. 

Verification. 

4 

+  4  + 

4  =  12. 

15.  The  sum  of  three  equal  numbers  is  24.  What  are  the 
numbers? 

16.  The  sum  of  three  equal  numbers  is  36.  What  are  the 
numbers? 

17.  The  sum  of  three  equal  numbers  is  54.  What  are  the 
numbers? 

Lesson  III. 

1.    What  number  is  that  which  added  to  three  times  itself 
will  make  48? 

Let  X  denote  the  number.  Then  3  a;  will  denote  three  times  the  num- 
ber;  and  X  plus  3  a;,  or  4  a;,  will  denote  the  sum,  which  is  48.  If  4  a;  is 
equ£il  to  48,  x  will  be  equal  to  48  divided  by  4,  which  is  12  :  therefore 
12  is  the  required  number.     Hence  we  write,  — 

Let  X  =  the  number. 

Then  3  a;  =  three  times  the  number, 


14  ELEMENTARY    ALGEBKA. 

and  a;  +  3a;  =  4a;  =  48,  the  sum. 

Then  x  =  —  =  12,  the  required  number. 

Veeification.        12  +  3  X  12  =  12  +  36  =  48. 

Note.  —  All  similar  questions  may  be  solved  by  the  same  form  of 
analysis. 

2.  What  number  added  to  four  times  itself  will  give  40? 

3.  What  number  added  to  five  times  itself  will  give  42? 

4.  What  number  added  to  six  times  itself  will  give  63? 

5.  What  number  added  to  seven  times  itself  will  give  88? 

6.  What  number  added  to  eight  times  itself  will  give  81  ? 

7.  What  number  added  to  nine  times  itself  will  give  100? 

8.  James  and  John  together  have  24  quills,  and  John  has 
three  times  as  many  as  James.     How  many  has  each  ? 

9.  William  and  Charles  have  64  marbles,  and  Charles  has 
seven  times  as  many  as  William.     How  many  has  each  ? 

10.  James  and  John  travel  96  miles,  and  James  travels 
eleven  times  as  far  as  John.     How  far  does  each  travel  ? 

11.  The  sum  of  the  ages  of  a  father  and  son  is  84  years, 
and  the  father  is  three  times  as  old  as  the  son.  What  is  the 
age  of  each  ? 

12.  There  are  two  numbers  of  which  the  greater  is  seven 
times  the  less,  and  their  sum  is  72.     What  are  the  numbers  ? 

13.  The  sum  of  four  equal  numbers  is  64.  What  are  the 
numbers  ? 

14.  The  sum  of  six  equal  numbers  is  54.  What  are  the 
numbers? 

15.  James  has  24  marbles.  He  loses  a  certain  number, 
and  then  gives  away  seven  times  as  many  as  he  loses,  which 
takes  all  he  has.     How  many  did  he  give  away?     Verify. 


INTRODUCTION.  15 

16.  William  has  36  cents,  and  divides  them  between  his 
two  brothers,  James  and  Charles,  giving  one  brother  eight 
times  as  many  as  the  other.  How  many  does  he  give  to 
each? 

17.  What  is  the  sum  of  x  and  3a;  f  Of  x  and  1x?  Oi  x 
and  hxf     Of  a;  and  Vlxf 

Lesson  IV. 

1.  If  1  apple  costs  1  cent,  what  will  a  number  of  apples 
denoted  by  x  cost  ? 

Since  1  apple  costs  1  cent,  and  since  x  denotes  any  number  of  apples, 
the  cost  of  X  apples  will  be  as  many  cents  as  there  are  apples ;  that  is, 
X  cents. 

2.  If  1  apple  costs  2  cents,  what  will  x  apples  cost? 

Since  1  apple  costs  2  cents,  and  since  x  denotes  the  number  of  apples, 
the  cost  will  be  twice  as  many  cents  as  there  are  apples;  that  is,  2x 
cents. 

3.  If  1  apple  costs  3  cents,  what  will  x  apples  cost? 

4.  If  1  lemon  costs  4  cents,  what  will  x  lemons  cost? 

5.  If  1  orange  costs  6  cents,  what  will  a  number  of  oranges 
denoted  by  x  cost  ? 

6.  Charles  bought  a  certain  number  of  lemons  at  2  cents 
apiece,  and  as  many  oranges  at  3  cents  apiece,  and  paid  in  all 
20  cents.     How  many  did  he  buy  of  each  ? 

Let  X  denote  the  number  of  lemons.  Then,  since  he  bought  as  many 
oranges  as  lemons,  it  will  also  denote  the  number  of  oranges.  Since  the 
lemons  were  2  cents  apiece,  2x  will  denote  the  cost  of  the  lemons ;  and 
since  the  oranges  were  3  cents  apiece,  3  a;  will  denote  the  cost  of  the 
oranges;  and  2x  plus  3a;,  or  5a;,  will  denote  the  cost  of  both,  which  is 
20  cents.  Now,  since  5x  cents  are  equal  to  20  cents,  a;  will  be  equal  to 
20  cents  divided  by  5  cents,  which  is  4:  hence  he  bought  4  of  each. 
Hence  we  write, — 


16  ELEMENTARY    ALGEBRA. 

Let  X  =  the  number  of  lemons,  or  orange?. 

Then  2  a;  =  the  cost  of  the  lemons, 

and  3  a;  =  the  cost  of  the  oranges. 

Hence  2a;  +  3a;  =  5a;  =  20  cents  =  the  cost  of  lemons  and  oranges. 

Hence  x  == =  4,  the  number  of  each. 

5  cents 

Verificatiox.     4x2  cents  =    8  cents,  cost  of  lemons. 
4x3  cents  =  12  cents,  cost  of  oranges. 

20  cents,  total  cost. 

7.  A  farmer  bought  a  certain  number  of  sheep  at  $4 
apiece,  and  an  equal  number  of  lambs  at  $1  apiece,  and  the 
whole  cost  $60.     How  many  did  he  buy  of  each  ? 

8.  Charles  bought  a  certain  number  of  apples  at  1  cent 
apiece,  and  an  equal  number  of  oranges  at  4  cents  apiece,  and 
paid  50  cents  in  all.     How  many  did  he  buy  of  each? 

9.  James  bought  an  equal  number  of  apples,  pears,  and 
lemons.  He  paid  1  cent  apiece  for  the  apples,  2  cents  apiece 
for  the  pears,  and  3  cents  apiece  for  the  lemons,  and  paid 
72  cents  in  all.     How  many  did  he  buy  of  each  ?     Verify. 

10.  A  farmer  bought  an  equal  number  of  sheep,  hogs,  and 
calves,  for  which  he  paid  $108.  He  paid  $3  apiece  for  the 
sheep,  $5  apiece  for  the  hogs,  and  $4  apiece  for  the  calves. 
How  many  did  he  buy  of  each? 

11.  A  farmer  sold  an  equal  number  of  ducks,  geese,  and 
turkeys,  for  which  he  received  90  shillings.  The  ducks 
brought  him  3  shillings  apiece,  the  geese  5,  and  the  turkeys  7. 
How  many  did  he  sell  of  each  sort? 

12.  A  tailor  bought,  for  $100,  two  pieces  of  cloth,  each  of 
which  contained  an  equal  number  of  yards.  For  one  piece 
he  paid  $3  a  yard,  and  for  the  other  $2  a  yard.  How  many 
yards  in  each  piece  ? 


INTRODUCTION.  17 

13.  The  sum  of  three  numbers  is  28.  The  second  is  twice 
the  first,  and  the  third  twice  the  second.  What  are  the 
numbers  ?     Verify. 

14.  The  sum  of  three  numbers  is  64.  The  second  is  three 
times  the  first,  and  the  third  four  times  the  second.  What 
are  the  numbers  ? 

Lesson  V. 

1.  If  1  yard  of  cloth  costs  x  dollars,  what  will  2  yards 

cost? 

Two  yards  of  cloth  will  cost  twice  as  much  as  one  yard :  therefore,  if 
1  yard  of  cloth  costs  x  dollars,  2  yards  will  cost  twice  x  dollars,  or 
2x  dollars. 

2.  If  1  yard  of  cloth  costs  x  dollars,  what  will  3  yards 
cost?     Why? 

3.  If  1  orange  costs  x  cents,  what  will  7  oranges  cost? 
Why?     8  oranges? 

4.  Charles  bought  3  lemons  and  4  oranges,  for  which  he 
paid  22  cents.  He  paid  twice  as  much  for  an  orange  as  for 
a  lemon.     What  was  the  price  of  each  ? 

Let  X  denote  the  price  of  a  lemon.  Then  2x  will  denote  the  price  of 
an  orange,  3  a;  will  denote  the  cost  of  3  lemons,  and  8  a;  the  cost  of 
4  oranges ;  hence  3  a;  plus  8  a;,  or  11a;,  will  denote  the  cost  of  the  lemons 
and  oranges,  which  is  22  cents.  If  11a;  is  equal  to  22  cents,  x  is  equal 
to  22  cents  divided  by  11,  which  is  2  cents :  therefore  the  price  of 
1  lemon  is  2  cents  ;  and  that  of  1  orange,  4  cents.     Hence  we  write,— 

Let  a;  =  the  price  of  I  lemon. 

Then  2  a;  =  the  price  of  1  orange, 

and  3a;  +  8a:  =  lla;  =  22  cents,  cost  of  lemons  and  oranges. 

Hence  a;  ==  -^ =  2  cents,  price  of  1  lemon  ; 

and  2  a;  =  2  X  2  cents  =  4  cents,  price  of  1  orange. 

D.  N.  E.  A.  —  2. 


18  ELEMENTARY    ALGEBRA. 

Verification.     3x2  cents  =    6  cents,  cost  of  lemons. 
4x4  cents  =  16  cents,  cost  of  oranges. 

22  cents,  total  cost. 

5.  James  bought  8  apples  and  3  oranges,  for  which  he 
•paid  20  cents.     He  paid  as  much  for  1  orange  as  for  4  apples. 

What  did  he  pay  for  one  of  each  ? 

6.  A  farmer  bought  3  calves  and  7  pigs,  for  which  he 
paid  $19.  He  paid  four  times  as  much  for  a  calf  as  for  a 
pig.     What  was  the  price  of  each  ? 

7.  James  bought  an  apple,  a  peach,  and  a  pear,  for  which 
he  paid  6  cents.  He  paid  twice  as  much  for  the  peach  as  for 
the  apple,  and  three  times  as  much  for  the  pear  as  for  the 
apple.     What  was  the  price  of  each  ? 

8.  William  bought  an  apple,  a  lemon,  and  an  orange,  for 
which  he  paid  24  cents.  He  paid  twice  as  much  for  the 
lemon  as  for  the  apple,  and  three  times  as  much  for  the 
orange  as  for  the  apple.     What  was  the  price  of  each  ? 

9.  A  farmer  sold  4  calves  and  5  cows,  for  which  he 
received  $120.  He  received  as  much  for  1  cow  as  for  4 
calves.     What  was  the  price  of  each  ? 

10.  Lucy  bought  3  pears  and  5  oranges,  for  which  she 
paid  26  cents,  giving  twice  as  much  for  each  orange  as  for 
each  pear.     What  was  the  price  of  each  ? 

11.  Ann  bought  2  skeins  of  silk,  3  pieces  of  tape,  and  a 
penknife,  for  which  she  paid  80  cents.  She  paid  the  same 
for  the  silk  as  for  the  tape,  and  as  much  for  the  penknife  as 
for  both.     What  was  the  cost  of  each  ? 

12.  James,  John,  and  Charles  are  to  divide  56  cents  among 
them  so  that  John  shall  have  twice  as  many  as  James,  and 
Charles  twice  as  many  as  John.     What  is  the  share  of  each? 


INTRODUCTION.  19 

13.  Put  54  apples  into  three  baskets  so  that  the  second 
shall  contain  twice  as  many  as  the  first,  and  the  third  as 
many  as  the  first  and  second.  How  many  will  there  be  in 
each  ? 

14.  Divide  60  into  four  such  parts  that  the  second  shall 
be  double  the  first,  the  third  double  the  second,  and  the 
fourth  double  the  third.     What  are  the  numbers? 

Lesson  VI. 

1.  If  2x-{-x  is  equal  to  Sx,  what  is  Sx  —  x  equal  to? 

Am.  (written)  Sx  —  x  =  2x. 

2.  What  is  4:X  —  x  equal  to? 

Ans.  (written)  Ax  —  x  =  Sx. 

3.  What  is  8a;  —  6:r  equal  to? 

Ans.  (written)  Sx—  6x  =  2x. 

4.  What  is    12a; ~  9a;   equal  to?  ^ns.  3a;. 

5.  What  is   15a;  — 7a;   equal  to?  ^ns.  8a;. 

6.  What  is   17a;  — 13a;   equal  to?  Ans.  ix. 

7.  Two  men,  who  are  30  miles  apart,  travel  towards  each 
other,  —  one  at  the  rate  of  2  miles  an  hour,  and  the  other  at 
the  rate  of  3  miles  an  hour.    How  long  before  they  will  meet? 

Let  X  denote  the  number  of  hours.  Then,  since  the  time  multiplied 
by  the  rate  will  give  the  distance,  2x  will  denote  the  distance  traveled 
by  the  first,  and  3  x  the  distance  traveled  by  the  second.  But  the  sum 
of  the  distances  is  30  miles :  hence  2a;  +  3a;  =  5a;  =  30  miles.  If  5 a;  is 
equal  to  30,  x  is  equal  to  30  divided  by  5,  which  is  6  :  hence  they  will 
meet  in  6  hours.     Hence  we  write,  — 

Let  X  =  the  time  in  hours. 

Then  2  a;  =  distance  traveled  by  1st, 

and  3.r  =  distance  traveled  by  2d. 


20  ELEMENTARY    ALGEBRA. 

By  the  conditions,    2a;  +  3a;  =  5a;  =  30  miles,  the  distance  apart. 

Hence  a;  =«  —  =  6  hours. 

5 

Verification.      2x6  miles  =  12  miles,  distance  traveled  by  1st. 
3x6  miles  =  18  miles,  distance  traveled  by  2d. 

30  miles,  whole  distance. 

8.  Two  persons  are  10  miles  apart,  and  are  traveling  in 
the  same  direction,  —  the  first  at  the  rate  of  3  miles  an  hour, 
and  the  second  at  the  rate  of  5  miles.  How  long  before  the 
second  will  overtake  the  first  ? 

Let  X  denote  the  time  in  hours.  Then  3ic  will  denote  the  distance 
traveled  by  the  first  in  x  hours,  and  5x  the  distance  traveled  by  the 
second.  But  when  the  second  overtakes  the  first,  he  will  have  traveled 
10  miles  more  than  the  first:  hence  5a;  —  3a:  =  2a;  =  10.  If  2a;  is  equal 
to  10,  x  is  equal  to  5:  hence  the  second  will  overtake  the  first  in  5 
hours.     Hence  we  write,  — 


Let 

X  =  the  time  in  hours. 

Then 

3  a;  =  the  distance  traveled  by  1st, 

and 

5  a;  =  the  distance  traveled  by  2d ; 

and 

5a;  -  3a;  =  2a;  =  10  hours, 

or 

X  =  —  =  5  hours. 

2 

Verification.    3x5  miles  =  15  miles,  distance  traveled  by  1st. 
5x5  miles  =  25  miles,  distance  traveled  by  2d. 

25  —  15  =  10  miles,  distance  apart. 

9.  A  cistern  holding  100  hogsheads  is  filled  by  two  pipes. 
One  discharges  8  hogsheads  a  minute,  and  the  other  12.  In 
what  time  will  they  fill  the  cistern  ? 

10.  A  cistern  holding  120  hogsheads  is  filled  by  3  pipes. 
The  first  discharges  4  hogsheads  in  a  minute,  the  second  7, 
and  the  third  1.     In  what  time  will  they  fill  the  cistern  ? 

11.  A  cistern  which  holds  90  hogsheads  is  filled  by  a  pipe 
which   discharges   10  hogsheads  a   minute ;    but  there   is   a 


INTRODUCTION.  21 

waste-pipe  which  loses  4  hogsheads  a  minute.     How  long  will 
it  take  to  fill  the  cistern  ? 

12.  Two  pieces  of  cloth  contain  each  an  equal  number  of 
yards.  The  first  cost  $3  a  yard,  and  the  second  $5,  and 
both  pieces  cost  $96.     How  many  yards  in  each  ? 

13.  Two  pieces  of  cloth  contain  each  an  equal  number  of 
yards.  The  first  cost  $7  a  yard,  and  the  second  $5.  The 
first  cost  $60  more  than  the  second.  How  many  yards  in 
each  piece  ? 

14.  John  bought  an  equal  number  of  oranges  and  lemons. 
The  oranges  cost  him  5  cents  apiece,  and  the  lemons  3,  and 
he  paid  56  cents  for  the  whole.  How  many  did  he  buy  of 
each  kind  ? 

15.  Charles  bought  an  equal  number  of  oranges  and  lem- 
ons. The  oranges  cost  him  5  cents  apiece,  and  the  lemons  3. 
He  paid  14  cents  more  for  the  oranges  than  for  the  lemons. 
How  many  did  he  buy  of  each  ? 

16.  Two  men  work  the  same  number  of  days.  The  one 
receives  $1  a  day,  and  the  other  $2.  At  the  end  of  the  time 
they  receive  $54.     How  long  did  they  work? 

Lesson  VII. 

1.    John  and  Charles  together  have  25  cents,  and  Charles 
has  5  more  than  John.     How  many  has  each  ? 

Let  X  denote  the  number  which  John  has.  Then  a;  +  5  will  denote 
the  number  which  Charles  has  ;  and  a;  +  a;  +  5,  or  2a;  +  5,  will  be  equal 
to  25,  the  number  they  both  have.  Since  2x  -\-b  equals  25,  2x  will  be 
equal  to  25  —  5,  or  20,  and  x  will  be  equal  to  20  divided  by  2,  or  10  : 
therefore  John  has  10  cents,  and  Charles  15.     Hence  we  write,  — 

Let  X  =  the  number  of  John's  cents. 

Then  a;  +  5  =  the  number  of  Charles's  cents, 

and  a;  +  :j:^  +  5  =  25,  the  number  they  both  have  ; 


2Z  KLEMENTARY    ALGEBPwA. 

or  2x  ^  5  ==  25, 

and  2  a;  =  25  -  5  -  20. 

Hence  a;  =  —         =10,  John's  number, 

2 

and  a:  +  5  =  10  +  5  =  15,  Charles's  number. 

John's.       Charles's. 

Veeification.      10     +     15    =  25,  the  sum. 

Charles's.       John's. 

15    —     10    =5,  the  difference. 

2.  James  and  John  have  30  marbles,  and  John  has  4 
more  than  James.     How  many  has  each  ? 

3.  William  bought  60  oranges  and  lemons.  There  were 
20  more  lemons  than  oranges.  How  many  were  there  of  each 
sort? 

4.  A  farmer  has  20  more  cows  than  calves.  In  all,  he 
has  36.     How  many  of  each  sort? 

5.  Lucy  has  28  pieces  of  money  in  her  purse,  composed 
of  cents  and  dimes.  The  cents  exceed  the  dimes  in  number 
by  16.     How  many  are  there  of  each  sort  ? 

6.  What  number  added  to  itself  and  to  9  will  make  29  ? 

7.  What  number  added  to  twice  itself  and  to  4  will  make 
25? 

8.  What  number  added  to  three  times  itself  and  to  12 
will  make  60  ? 

9.  John  has  five  times  as  many  marbles  as  Charles;  and 
what  they  both  have,  added  to  14,  makes  44.  How  many 
has  each  ? 

10.  There  are  three  numbers,  of  which  the  second  is  twice 
the  first,  and  the  third  twice  the  second  ;  and  when  9  is  added 
to  the  sum,  the  result  is  30.     What  are  the  numbers? 


INTRODUCTIOX.  'S6 

11.  Divide  17  into  two  such  parts  that  the  second  shall  be 
two  more  than  double  the  first.     What  are  the  parts  ? 

12.  Divide  40  into  three  such  parts  that  the  second  shall 
be  twice  the  first,  and  the  third  exceed  six  times  the  first  by  4. 
What  are  the  parts  ? 

13.  Charles  has  twice  as  many  cents  as  James,  and  John 
has  twice  as  many  as  Charles.  If  7  be  added  to  what  they 
all  have,  the  sum  will  be  28.     How  many  has  each  ? 

14.  Divide  16  into  three  such  parts  that  the  second  shall 
be  three  times  the  first ;  the  third  twice  the  second,  and  5 
over.     What  are  the  numbers? 

15.  An  orchard  contains  three  kinds  of  trees,  —  apple, 
pear,  and  cherry.  There  are  four  times  as  many  pear  as 
apple  trees ;  twice  as  many  cherry  as  pear  trees ;  and  if  14  be 
added,  the  number  will  be  40.    How  many  are  there  of  each  ? 

Lesson  VIII. 

1.  John,  after  giving  away  5  marbles,  had  12  left.     How 
many  had  he  at  first  ? 

Let  X  denote  the  number.  Then  x  —  b  will  denote  what  he  had  left, 
which  was  equal  to  12.  Since  x  diminislffed  by  5  is  equal  to  12,  x  will 
be  equal  to  12  increased  by  5 ;  that  is,  to  17 :  therefore  he  had  17  marbles. 
Hence  we  write,  — 

Let  X  =  the  number  he  had  at  first. 

Then  x  -5  =  12,  what  he  had  left ; 

and  a;  =  12  +  5  =  17,  what  he  first  had. 

Verificatioit.     17  -  5  =  12,  what  were  left. 

2.  Charles  lost  6  marbles,  and  has  9  left.      How  many 
had  he  at  first? 

3.  William  gave  15  cents  to  John,  and  had  9  left.     How 
many  had  he  at  first  ? 


24  ELEMENTARY    ALGEBRA. 

4.  Ann  plucked  8  buds   from   her   rosebush,  and   there 
were  19  left.     How  many  were  there  at  first  ? 

5.  William  took  27  cents  from  his  purse,  and  there  were 
13  left.     How  many  were  there  at  first? 

6.  The  sum  of  two  numbers  is  14,  and  their  difference 
is  2.     What  are  the  numbers  ? 

The  difference  of  two  numbers,  added  to  the  less,  will  give  the  greater. 
Let  X  denote  the  less  number.  Then  x  +  2  will  denote  the  greater ;  and 
a;  +  a;  +  2  will  denote  their  sum,  which  is  14.  Then  2a;  +  2  equals  14 ; 
and  2a;  equals  14  —  2,  or  12:  hence  x  equals  12  divided  by  2,  or  6: 
therefore  the  numbers  are  6  and  8. 

Veeification.  8  +  6  =  14,  their  sum  ; 

and     8  —  6  =    2,  their  difference. 

7.  The  sum  of  two  numbers  is  18,  and  their  difference  6. 
What  are  the  numbers  ? 

8.  James  and  John  have  26  marbles,  and  James  has  4 
more  than  John.     How  many  has  each  ? 

9.  Jane  and  Lucy  have  16  books,  and  Lucy  has  8  more 
than  Jane.     How  many  has  each  ? 

10.  William  bought  an  equal  number  of  oranges  and 
lemons.  Charles  took  5  lemons,  after  which  William  had 
but  25  of  both  sorts.     How  many  did  he  buy  of  each? 

11.  Mary  has  an  equal  number  of  roses  on  each  of  two 
bushes.  If  she  takes  4  from  one  bush,  there  will  remain  24 
on  both.     How  many  on  each  at  first  ? 

12.  The  sum  of  two  numbers  is  20,  and  their  difference 
is  6.     What  are  the  numbers? 

If  X  denotes  the  greater  number,  x  —  ^  will  denote  the  less,  and 
x-\-x  —  Qt  will  be  equal  to  20:  hence  2a;  equals  20  +  6,  or  26;  and  x 
equals  26  divided  by  2,  equals  13 :  therefore  the  numbers  are  13  and  7. 
Hence  we  write,  — 


INTRODUCTION.  25 

Let  X  =  the  greater. 

Then  aj  _  6  =-  the  less, 

and  a;  +  a;  —  6  =  20,  their  sum. 
Hence  2a;  =  20  +  6  =  26; 

or  ar  =  —  =  13,  and  13-6-7. 

2 

Verification.     13  +  7  =  20,  and  13  -  7  =  6. 

13.  The  sum  of  the  ages  of  a  father  and  son  is  60  years, 
and  their  difference  is  just  half  that  number.  What  are  their 
ages? 

14.  The  sum  of  two  numbers  is  23,  and  the  larger  lacks  1 
of  being  seven  times  the  smaller.     What  are  the  numbers  ? 

15.  The  sum  of  two  numbers  is  50.  The  larger  is  equal 
to  ten  times  the  less,  minus  5.     What  are  the  numbers  ? 

16.  John  has  a  certain  number  of  oranges,  and  Charles  has 
four  times  as  many,  less  7.  Together  they  have  53.  How 
many  has  each  ? 

17.  An  orchard  contains  a  certain  number  of  apple  trees, 
and  three  times  as  many  cherry  trees,  less  6.  The  whole 
number  is  30.     How  many  of  each  sort  ? 

Lesson  IX. 

1.  li  X  denotes  any  number,  and  1  be  added  to  it,  what 
will  denote  the  sum  ?  Ans.  x+\. 

2.  If  2  be  added  to  x,  what  will  denote  the  sum  ?     If  3 
be  added,  what?     If  4  be  added?  etc. 

3.  If  to  John's  marbles   1   marble  be  added,  twice   his 
number  will  be  equal  to  10.     How  many  had  he  ? 

Let  X  denote  the  number.  Then  a;  +  1  will  denote  the  number  after 
1  is  added;  and  twice  this  number,  or  2 a?  -f  2,  will  be  equal  to  10.     If 

i 


26  ELEMENTARY    ALGEBRA. 

2.r  +  2  is  equal  to  10,  2x  will  be  equal  to  10  —  2,  or  8;  or  a;  will  be 
equal  to  4.     Hence  we  write,  — 

Let  X  =  the  number  of  John's  marbles. 

Then  a;  +  1  =  the  number  after  1  is  added, 

and  2(a;4-l)  =  2a;  +  2  =  10. 
Hence  2  a;  =10 -2, 

8  A 

or  X  =  -  =  4. 

2 

Verification.  2  (4  +  1)  =  2x5  =  10. 

4.  Write  x  +  2  multiplied  by  3.  Am.  3  {x  +  2). 
What  is  the  product?  Ans.  3:r  +  6. 

5.  Write  a;  +  4  multiplied  by  5.  Ans.  5  (a;  +  4). 
What  is  the  product?  Ans.  bx  +  20, 

6.  Write  x  +  2t  multiplied  by  4.  Ans.  4  (a;  -f  3). 
What  is  the  product  ?  Ans.  4  a; +12. 

7.  Lucy  has  a  certain  number  of  books.  Her  father  gives 
her  two  more,  when  twice  her  number  is  equal  to  14.  How 
many  has  she  ? 

8.  Jane  has  a  certain  number  of  roses  in  blossom.  Two 
more  bloom,  and  then  three  times  the  number  is  equal  to  15. 
How  many  were  in  blossom  at  first? 

9.  Jane  has  a  certain  number  of  handkerchiefs,  and  buys 
four  more,  when  five  times  her  number  is  equal  to  45.  How 
many  had  she  at  first  ? 

10.  John  has  one  apple  more  than  Charles,  and  three 
times  John's  added  to  what  Charles  has  make  15.  How 
many  has  each? 

Let  X  denote  Charles's  apples.  Then  a;  +  1  will  denote  John's ;  and 
a;  +  1  multiplied  by  3,  added  to  a;,  or  3a:  +  3  +  a;,  will  be  equal  to  15, 
what  they  both  had:  hence  4 a; +  3  equals  15;  and  4a;  equals  15  —  3, 
or  12  ;  and  a:  =  4.     Write  and  verify. 


INTRODUCTION.  27 

11.  James  lias  two  marbles  more  than  William,  and  twice 
his  marbles  plus  twice  William's  are  equal  to  16.  How  many 
has  each? 

12.  Divide  20  into  two  such  parts  that  one  part  shall  ex- 
ceed the  other  by  4. 

13.  A  fruit-bast et  contains  apples,  pears,  and  peaches. 
There  are  2  more  pears  than  apples,  and  twice  as  many 
peaches  as  pears.  There  are  22  in  all.  How  many  of  each 
sort  ? 

14.  What  isthe  sumof  rr  +  3:?;  +  2(:r+l)? 

15.  What  is  thesumof  2(rr-f  l)  +  l(:i'+l)  +  a:? 

16.  What  is  the  sum  of  x  +  5(x  +  S)? 

17.  The  sum  of  two  numbers  is  11,  and  the  second  is  equal 
to  twice  the  first  plus  2.     What  are  the  numbers  ? 

18.  John  bought  3  apples,  3  lemons,  and  3  oranges,  for 
which  he  paid  27  cents.  He  paid  one  cent  more  for  a  lemon 
than  for  an  apple,  and  1  cent  more  for  an  orange  than  for  a 
lemon.     What  did  he  pay  for  each  ? 

19.  Lucy,  Mary,  and  Ann  have  15  cents.  Mary  has  1 
more  than  Lucy,  and  Ann  twice  as  many  as  Mary.  How 
many  has  each  ? 

Lesson  X. 

1.  li  X  denote  any  number,  and  1  be  subtracted  from  it, 
what  will  denote  the  difference  ?  Ans.  x  —  1. 

If  2  be  subtracted,  what  will  denote  the  difference?     If  3, 
be  subtracted  ?  4  ?  etc. 

2.  John  has  a  certain  number  of  marbles.  If  1  be  taken 
away,  twice  the  remainder  will  be  equal  to  12.  How  many 
has  he  ? 


28  ELEMENTARY    ALGEBRA. 

Let  X  denote  the  number.  Then  x  —  1  will  denote  the  number  after 
1  is  taken  away ;  and  twice  this  number,  or  2  (a;  —  1)  =  2  a;  —  2,  will  be 
equal  to  12.  \i  2x  diminished  by  2  is  equal  to  12,  2a;  is  equal  to  12  -|-  2, 
or  14  :  therefore  x  equals  14  divided  by  2,  or  7.     Hence  we  write,  — 

Let  X  =  the  number. 

Then  a;  —  1  =  the  number  which  remained, 

and  2(a;-l)  =  2a;-2  =  12. 

Hence  2a;  =  12  +  2,  or  14  ; 

and  x  =  —  =  1. 

2 

Verification.  2  (7  -  1)  =  14  -  2  =  12 ; 

also  2(7-1)=    2x6  =  12. 

3.  Write  3  times  x  —  1.  Arts.  3  (rr  —  1). 
What  is  the  product  equal  to  ?  Arts.  3  a;  —  3. 

4.  Write  4  times  x  —  2.  Ayis.  ^{x  —  2). 
What  is  the  product  equal  to?  Ans.  4a;  —  8. 

6.   Write  5  times  a;  —  5.  Ans.  b{x  —  5). 

What  is  the  product  equal  to  ?  Ans.  5  a;  —  25. 

6.  If  X  denotes  a  certain  number,  will  x  —  1  denote  a 
greater,  or  less,  number  ?     How  much  less  ? 

7.  If  a;  —  1  is  equal  to  4,  what  will  x  be  equal  to  ? 

Ans.  4  +  1,  or  5. 

8.  If  a;  —  2  is  equal  to  6,  what  is  x  equal  to? 

9.  James  and  John  together  have  20  oranges.     John  has 
6  less  than  James.     How  many  has  each  ? 

10.  A  grocer  sold  12  pounds  of  tea  and  coffee.  If  the  tea 
be  diminished  by  3  pounds,  and  the  remainder  multiplied  by 
2,  the  product  is  the  number  of  pounds  of  coffee.  How  many 
pounds  of  each? 

11.  Ann  has  a  certain  number  of  oranges.  Jane  has  1 
less,  and  twice  her  number  added  to  Ann's  make  13.  How 
many  has  each  ? 


INTRODUCTION.  29 

Let  X  denote  the  number  of  oranges  which  Ann  has.  Then  x  —  1 
will  denote  the  number  Jane  has;  and  x-[-2x  —  2,  or  3a;  — 2,  will 
denote  the  number  both  have,  which  is  13.  If  3ir  — 2  equals  13,  3  a; 
will  be  equal  to  13  +  2,  or  15;  and  if  3  a;  is  equal  to  15,  x  will  be  equal 
to  15  divided  by  3,  which  is  5  :  therefore  Ann  has  5  oranges,  and  Jane  4. 
Hence  we  write,  — 

Let  X  =  the  number  Ann  has. 

Then  a;  —  1  =  the  number  Jane  has, 

and  2  (a-  —  1)  =  2a;  —  2  =  twice  what  Jane  has, 

also  a;  +  2a;-2  =  3a;-2  =  13. 

Hence  3a;  =  13  +  2  =  15; 

or  X  =  —  =  5. 

3 

Verification.      5-4  =  1;  and  2  x  4  +  5  =  13. 

12.  Charles  and  John  have  20  cents,  and  John  has  6  less 
than  Charles.     How  many  has  each? 

13.  James  has  twice  as  many  oranges  as  lemons  in  his 
basket,  and  if  5  be  taken  from  the  whole  number,  19  will 
remain.     How  many  had  he  of  each? 

14.  A  basket  contains  apples,  peaches,  and  pears,  29  in  all. 
If  1  be  taken  from  the  number  of  apples,  the  remainder  will 
denote  the  number  of  peaches,  and  twice  that  remainder  will 
denote  the  number  of  pears.  How  many  are  there  of  each 
sort  ? 

15.  If  2a:  —  5  equals  15,  w^hat  is  the  value  of  x  f 

16.  If  4a;  —  5  is  equal  to  11,  what  is  the  value  of  x  f 

17.  If  5  a;  —  12  is  equal  to  18,  what  is  the  value  of  x  f 

18.  The  sum  of  two  numbers  is  32,  and  the  greater  exceeds 
the  less  by  8.     What  are  the  numbers  ? 

19.  The  sum  of  two  numbers  is  9.  If  the  greater  number 
be  diminished  by  5,  and  the  remainder  multiplied  by  3,  the 
product  will  be  the  less  number.     What  are  the  numbers? 


3D  ELEMENTARY    ALGEBRA. 

20.  There  are  three  numbers  such  that  1  taken  from  the 
first  will  give  the  second,  the  second  multiplied  by  3  will 
give  the  third,  and  their  sum  is  equal  to  26.  What  are  the 
numbers  ? 

21.  John  and  Charles  together  have  just  81  oranges.  If  1 
be  taken  from  John's,  and  the  remainder  be  multiplied  by  5, 
the  product  will  be  equal  to  Charles's  number.  How  many 
has  each  ? 

22.  A  basket  is  filled  with  apples,  lemons,  and  oranges;  in 
all,  26.  The  number  of  lemons  exceed  the  number  of  apples 
by  2,  and  the  number  of  oranges  is  double  that  of  the  lemons. 
How  many  are  there  of  each  ? 

Lesson  XL 

1.  John  has  a  certain  number  of  apples,  the  half  of  which 
is  equal  to  10.     How  many  has  he  ? 

Let  X  denote  the  number  of  apples.  Then  x  divided  by  2  is  equal  to 
10.  If  one  half  of  x  is  equal  to  10,  twice  one  half  of  x,  or  x,  is  equal 
to  twice  10,  which  is  20 :  hence  x  is  equal  to  20. 

Note.  —  A  similar  analysis  is  applicable  to  any  one  of  the  fractional 
units.     Let  each  question  be  solved  according  to  the  analysis. 

2.  John  has  a  certain  number  of  oranges,  and  one  third 
of  his  number  is  15.     How  many  has  he? 

3.  If  one  fifth  of  a  number  is  6,  what  is  the  number? 

4.  If  one  twelfth  of  a  number  is  9,  what  is  the  number? 

6.    What  number  added  to  one  half  of  itself  will  give  a 
sum  equal  to  12? 

Denote  the  number  by  x.  Then  x  plus  one  half  of  x  equals  12.  But 
X  plus  one  half  of  x  equals  three  halves  of  x:  hence  three  halves  of  x 
equal  12.  If  three  halves  of  x  equal  12,  one  half  of  x  equals  one  third 
of  12,  or  4.  If  one  half  of  x  equals  4,  x  equals  twice  4,  or  8:  therefore 
X  equals  8.     Hence  we  write,  — 


''"\i^  .  >■  'jr.'       . 

INTRODUCTION. 

Let 

X  =  the  number. 

Then 

x-\-^x=^^x==\2, 
2        2 

aild 

-x  =  i,  or  x  =  8. 
2 

Veri 

FICATION. 

8 +5  =  8+4=  12. 

31 


6.  What  number  added  to  one  third  of  itself  will  give  a 
sum  equal  to  12  ? 

7.  What  number  added  to  one  fourth  of  itself  will  give  a 
sum  equal  to  20? 

8.  What  number  added  to  a  fifth  of  itself  will  make  24  ? 

9.  What  number  diminished   by  one   half  of  itself  will 
leave  4?     Why? 

10.  What  number  diminished  by  one  third  of  itself  will 
leave  6? 

11.  James  gave  one   seventh  of  his  marbles  to  William, 
and  then  had  24  left.     How  many  had  he  at  first  ?      « 

12.  What  number  added  to  two  thirds  of  itself  will  give  a 
sum  equal  to  20  ? 

13.  ^yhat  number  diminished  by  three  fourths  of  itself  will 
leave  9  ? 

14.  What   number   added  to  five   sevenths   of  itself  will 
make  24  ? 

15.  What  number  diminished    by  seven  eighths  of  itself 
will  leave  4  ? 

16.  What   number   added    to   eight   ninths   of  itself  will 
make  34  ? 

17.  What  number  diminished  by  eleven  twelfths  of  itself 
will  leave  5? 


32  ELEMENTARY    ALGEBRA. 

18.  Margaret  gave  nine  tenths  of  her  apples  to  her  sister, 
and  then  had  6  left.     How  many  had  she  at  first  ? 

19.  What  number  added  to  3  times  one  ninth  of  itself  will 
give  72? 

20.  Henry  had  a  certain  number  of  cents.     He  lost  one 
third  of  them,  and  had  15  left.     How  many  had  he  at  first  ? 


CHAPTER  I. 
DEFINITIONS. 

1.  Quantity  is  anything  which  can   be  increased,  dimin- 
ished, and  measured  ;  as  number,  distance,  weight,  time,  etc. 

To  measure  a  thing  is  to  find  how  many  times  it  contains 
some  other  thing  of  the  same  kind,  taken  as  a  standard.  The 
assumed  standard  is  called  the  unit  of  measure. 

2.  Mathematics  is  the  science  which  treats  of  the  measure- 
ment, properties,  and  relations  of  quantities. 

In  pure  mathematics  there  are  but  eight  kinds  of  quantity, 
and  consequently  but  eight  kinds  of  units ;  viz.,  units  of 
number,  units  of  currency,  units  of  length,  units  of  surface, 
units  of  volume,  units  of  weight,  units  of  time,  and  units  of 
angular  measure. 

3.  Algebra  is  a  branch  of  mathematics  in  which  the  quan- 
tities considered  are  represented  by  letters,  and  the  operations 
to  be  performed  are  indicated  by  signs. 

4.  The  quantities  employed  in  algebra  are  of  two  kinds,  — 
known  and  unknown, 

Known  quantities  are  those  whose  values  are  given.  They 
are  generally  represented  by  the  leading  letters  of  the  alpha- 
bet ;  as,  a,  h,  c,  etc. 

Unknown  quantities  are  those  whose  values  are  required. 
They  are  generally  represented  by  the  final  letters  of  the 
alphabet ;  as,  x,  y,  z,  etc. 

When  an  unknown  quantity  becomes  known,  it  is  often 
denoted  by  the  same  letter,  with  one  or  more  accents;   as, 

D.  N.  E.  A,  —  8.  33 


34  ELEMENTARY    ALGEBRA. 

x\  x'\  x''\     These  symbol"fe  are  read,  "  x  prime,"  "  x  second," 
''  X  third." 

5.  The  sign  of  addition  (+)  is  called  plus,  When  placed 
between  two  quantities,  it  indicates  that  the  second  is  to  be 
added  to  the  first.  Thus,  a  +  b  is  read  *'  a  plus  3,"  and  indi- 
cates that  b  is  to  be  added  to  a.  If  no  sign  is  written,  the 
sign  +  is  understood. 

The  sign  +  is  sometimes  called  the  positive  sign;  and  the 
quantities  before  which  it  is  written  are  called  positive  quan- 
tities, or  additive  quantities. 

6.  The  sign  of  subtraction  (— )  is  called  minus.  When 
placed  between  two  quantities,  it  indicates  that  the  second  is 
to  be  subtracted  from  the  first.  Thus,  the  expression  c  —  d, 
read  "  c  minus  c?,"  indicates  that  d  is  to  be  subtracted  from  c. 
If  a  stands  for  6,. and  d  for  4,  then  a  —  c?  is  equal  to  6  —  4, 
which  is  equal  to  2. 

The  sign  —  is  sometimes  called  the  negative  sign;  and  the 
quantities  before  which  it  is  written  are  called  negative  quan- 
tities, or  subtractive  quantities. 

7.  The  sign  of  multiplication| (x)  is  read  "multiplied  by," 
or  "  multiplied  into."  When  fplaced  between  two  quantities, 
it  indicates  that  the  first  is  to  be  multipliedjby  the  second. 
Thus,  a  X  5  indicates  that  a  i§  to  be  multiplied  by  b.  If  a 
stands  for  7,  and  b  for  5,  then  a  X  5  is  equal  to  7  X  5,  which 
is  equal  to  35. 

The  multiplication  of  quantities  is  also.-indicated  by  simply^ii^* 
writing  the  letters  one  after  the  oth^r,   and  sometimes  by 
placing  a  point  between  them.     Thus,  a  X  b  signifies  the  same 
thing  as  a5  or  as  a .  5 ;  axbx  c  signifies  the  same  thing  as 
abc  or  as  a .  5  .  c. 

8.  A  factor  is  any  one  of  the  multipliers  of  a  product. 
Factors  are  of  two  kinds,  —  numeral  and  literal.     Thus,  in  the 


/ 


DEFINITIONS.  35 

expression  babe  there  are  four  factors,  —  the  numeral  factor  5, 
and  the  three  literal  factors,  a,  b,  and  c. 

9.  The  sign  of  division  (-^)  is  read  "  divided  by."  When 
written  between  two  quantities,  it  indicates  that  the  first  is 
to  be  divided  by  the  second. 

There   are   three   signs   used   to   denote   division.      Thus, 

a-^b,  -,  a\b,  denote  that  a  is  to  be  divided  by  b. 

b       ~ 

10.  The  sign  of  equality  (— )  is  read  "■  equal  to."  When 
written  between  two  quantities,  it  indicates  that  they  are 
equal  to  each  other.  Thus,  the  expression  a  +  b  =  c  indicates 
that  the  sum  of  a  and  b  is  equal  to  c.  If  a  stands  for  3,  and 
b  for  5,  c  will  be  equal  to  8. 

11.  The  sign  of  inequality  (>  or  <)  is  read  "  greater  than  " 
or  "  less  than."  When  placed  between  two  quantities,  it  indi- 
cates that  they  are  unequal,  the  greater  one  being  placed  at 
the  opening  of  the  sign.  Thus,  the  expression  a>b  indicates 
that  a  is  greater  than  b,  and  the  expression  c  <  c?  indicates 
that  c  is  less  than  d, 

12.  The  sign  .*.  means  "  therefore,"  or  ''  consequently." 

13.  A  coefficient  is  a  number  written  before  a  quantity,  to 
show  how  many  times  the  quantity  is  taken  additively.  Thus, 
in  the  expression  a-\-a-{-a-\-a-\-a  =  ba,  5  is  the  coefficient 
of  a. 

A  coefficient  may  be  denoted  either  by  a  number  or  by  a 
letter.  Thus,  bx  indicates  that  x  is  taken  5  times,  and  ax 
indicates  that  x  is  taken  a  times.  If  no  coefficient  is  written, 
the  coefficient  1  is  understood.     Thus,  a  is  the  same  as  la. 

14.  An  exponent  is  a  number  written  at  the  right  and  a 
little  above  a  quantity,  to  indicate  how  many  times  the  quan- 
tity is  taken  as  a  factor.     Thus, 


36  ELEMENTARY    ALGEBRA. 

a  X  a  is  written  a'V 
aXaXa         "         a^ 
aX  aXaXa         "         a\  etc. 

and  the  2,  3,  and  4  are  exponents.  The  expressions  are  read, 
**  a  square,"  "a  cube"  or  "a  third,"  "a  fourth;"  and  if  we 
have  a"*,  in  which  a  enters  m  times  as  a  factor,  it  is  read  "  a 
to  the  mth,"  or  simply  "a  mth."  The  exponent  1  is  generally- 
omitted.  Thus,  a^  is  the  same  as  a,  each  denoting  that  a 
enters  but  once  as  a  factor. 

15.  A  power  is  a  product  which  arises  from  the  multiplica- 
tion of  equal  factors.     Thus, 

aX  a^^G^  =  the  square,  or  second  power,  of  a, 
a  X  aX  a=^  a^  =  the  cube,  or  third  power,  of  a. 
aXaXaXa  =  d^=  the  fourth  power  of  a. 
aX  aX  .  .  .  .  =  a"*  =  the  mth  power  of  a, 

16.  A  root  of  a  quantity  is  one  of  the  equal  factors.  The 
radical  sign  (^  ),  when  placed  over  a  quantity,  indicates 
that  a  root  of  that  quantity  is  to  be  extracted.  The  root  is 
indicated  by  a  number  written  over  the  radical  sign,  called  an 
index.     When  the  index  is  2,  it  is  generally  omitted.     Thus, 

Va,  or  Va,  indicates  the  square  root  of  a. 
Va   indicates  the  cube  root  of  a. 
Va   indicates  the  fourth  root  of  a. 
Va   indicates  the  mth  root  of  a. 

17.  An  algebraic  expression  is  a  quantity  written  in  alge- 
^braic   language.      Thus,    3  a   is   the    algebraic    expression    of 

three  times  the  number  denoted  by  a;  6a^  that  of  five  times 
the  square  of  a;  7aW,  that  of  seven  times  the  cube  of  a 
multiplied  by  the  square  of  ^;  3  a  — 56,  that  of  the  diflfer- 


DEFINITIONS.  6i 

ence  between  three  times  a  and  five  times  b ;  and  2  0^  — Sab 
-f  4^^  that  of  twice  the  square  of  a,  diminished  by  three 
times  the  product  of  a  by  b,  augmented  by  four  times  the 
square  of  b. 

18.  A  term  is  an  algebraic  expression  that  can  be  written 
without  the  aid  of  either  of  the  signs  +  or  — .  Thus,  3  a, 
2ab,  ba^b\  are  terms. 

A  term  may  be  preceded  by  either  of  the  signs  +  or  — . 
In  this  case  the  sign  is  called  the  sign  of  the  term,  and  is  used 
to  show  the  sense  in  which  the  term  is  taken.  Thus  +3a 
shows  that  3  a  is  taken  positively,  and  —baV  shows  that 
5  a^  b'^  is  taken  negatively. 

19.  The  degree  of  a  term  is  the  number  of  its  literal  factors. 
Thus,  3  a  is  a  term  of  the  first  degree,  because  it  contains  but 
one  literal  factor  ;  5a'^  is  a  term  of  the  second  degree,  because 
it  contains  two  literal  factors  (the  factors  in  this  case  are 
equal);  lo^b  is  a  term  of  the  fourth  degree,  because  it  con- 
tains four  literal  factors. 

The  degree  of  a  term  is  determined  by  the  sum  of  the 
exponents  of  all  its  letters. 

20.  A  monomial  is  a  single  term,  unconnected  with  any 
other  by  the  signs  +  oi*  —  •     Thus,  3a^,  3^^  a,  are  monomials. 

21.  A  polynomial  is  a  collection  of  terms  connected  by  the 
signs  +  or  — ;  as,  3a  —  5,  or  2a^  —  3Z)  +  4 b^. 

22.  A  binomial  is  a  polynomial  of  two  terms ;   as,  a  +  ^, 

23.  A  trinomial  is  a  polynomial  of  three  terms ;  as, 
ahc  —  a^  +  ^^  <^^  —  9^  —J- 

24.  Homogeneous  terms  are  those  which  are  of  the  same 
degree.  Thus,  the  terms  abc,  —  a^,  +  &,  are  homogeneous ; 
as  are  the  terms  ab,  —  gh. 


38  ELEMENTARY   ALGEBRA. 

25.  A  polynomial  is  homogeneous  when  all  its  terms  are 
homogeneous.  Thus,  the  polynomial  ahc  —  a^  +  c^  is  homo- 
geneous, but  the  polynomial  ab  —  gh  —f  is  not  homogeneous. 

26.  Similar  terms  are  those  which  have  a  common  unit ; 
that  is,  have  a  common  literal  part.  Thus,  1  ab -\-Zab  —  2ab 
are  similar  terms,  and  so  also  are  ^a^b"^ —  2a}b'^ —  ^a^b"^  ]  but 
the  terms  of  the  first  polynomial  and  of  the  last  are  not 
similar. 

27.  The  vinculum,  ;    the  bar,   | ;    the  parentheses,   (  ) ; 

and  the  brackets,  [  ],  — are  each  used  to  connect  two  or  more 
terms  which  are  to  be  operated  upon  as  a  whole.  Thus,  each 
of  the  expressions 


{a  +  b  +  c)xx,       [a  +  b  +  c\Xx, 


a  +  b  +  cXx,     +b 

+  c 

indicates  that  the  sum  of  a,  b,  and  e,  is  to  be  multiplied  by  x, 

28.  The  reciprocal  of  a  quantity  is  1  divided  by  that  quan- 
tity.    Thus, 

-,    ,    -,    are  the  reciprocals  of   a,    a  +  b,   — 

a    a+b     d  c 

29.  The  numerical  value  of  an  algebraic  expression  is  the 
result  obtained  by  assigning  a  numerical  value  to  each  letter, 
and  then  performing  the  operations  indicated.  Thus,  the 
numerical  value  of  the  expression  ab  +  bc  +  d,  when  a=l, 
b:=2,  c-=3,  and  d  =  ^,  is  1x2  +  2x3  +  4  =  12,  by  per- 
forming the  indicated  operations. 

Exercises  in  Writing  Algebraic  Expressions. 

1.  Write  a  added  to  b.  Ans.  a-{-  b. 

2.  Write  b  subtracted  from  a.  Am.  a—b. 


ALGEBRAIC   EXPRESSIONS.  6U 

Write  the  following  :  — 

3.  Six  times  the  square  of  a,  minus  twice  the  square  of  b. 

4.  Six  times  a  multiplied  by  Z>,  diminished  by  5  times  c 
cube  multiplied  by  d. 

5.  Nine  times  a,  multiplied  by  c  plus  c?,  diminished  by 

8  times  b  multiplied  by  d  cube. 

6.  Five  times  a  minus  b,  plus  6  times  a  cube  into  b  cube. 

7.  Eight  times  a  cube  into  d  fourth,  into  c  fourth,  plus 

9  times  c  cube  into  d  fifth,  minus  6  times  a  into  b,  into  c 
square. 

8.  Fourteen  times  a  plus  b,  multiplied  by  a  minus  5,  plus 
5  times  a,  into  c  plus  d, 

9.  Six  times  a,  into  c  plus  d,  minus  5  times  b,  into  a  plus 
c,  minus  4  times  a  cube  b  square. 

10.  Write  a,  multiplied  by  c  plus  d,  plus/ minus  g. 

11.  Write  a  divided  hj  b -\- c  in  three  ways. 

12.  Write  a  —  b  divided  by  a  +  b. 

13.  Write  a  polynomial  of  three  terms,  of  four  terms,  of 
five,  of  six. 

14.  Write  a  homogeneous  binomial  of  the  first  degree,  of 
the  second,  of  the  third,  fourth,  fifth,  sixth. 

15.  Write  a  homogeneous  trinomial  of  the  first  degree,  with 
its  second  and  third  terms  negative ;  of  the  second  degree ; 
of  the  third  ;  of  the  fourth. 

16.  Write  in  the  same  column,  on  the  slate  or  blackboard, 
a  monomial ;  a  binomial ;  a  trinomial ;  a  polynomial  of  four 
terms,  of  five  terms,  of  six  terms,  and  of  seven  terms ;  and  all 
of  the  same  degree. 


40  elementary  algebra. 

Exercises  in  Interpretation  of  Algebraic  Language. 

Find  the  numerical  values  of  the  following  expressions  when 
a  =  l,  b  =  2,  c  =  3,  and  <i  =  4:  — 

1.  ab  +  be. 

2.  a  +  bc  +  d. 

3.  ad-\-b  —  c. 

4.  ab  +  bc  —  d. 

5.  (a  +  b)c^-d. 

6.  (a  +  b)(d-b). 

7.  (ab -\- ad)  c  +  d. 

8.  (ab -\- c)(ad  —  a). 

9.  3a'b^-2(a  +  d+l), 

10.   9±Sx(a  +  d). 

^^     a'  +  b'-^  c'  ^  a^  +  P  +  c^'-d 

7  •  2  * 

1A» X 


Ans. 

8. 

Arts. 

11. 

Ans. 

3. 

Ans. 

4. 

Ans. 

23. 

Ans. 

6. 

Ans. 

22. 

Ans. 

15. 

Ans. 

0. 

Ans. 

10. 

Ans. 

32. 

Ans. 

4. 

6  33 

Find  the  numerical  values  of  the  following  expressions  when 
a  =  4,  b  =  S,  c  =  2,  and  d=l:  — 

13.  ^-^-\-c-d.  Ans.    2. 

14.  5^|-^\  ^ns.l5. 
16.    [(a'b  +  l)d]^(a'b  +  d).  Ans.    1. 

16.  4  /'a&c  -  ^"j  X  (30  e»  -  ai»(Z»).  ^ns.  11088. 

17.  ^±141+  ^+  4«\+^'--<  Ans.  Hf 

18.  ^^(^  +  f+^)-^  +  -|-XaWc^'.  ^n..3465. 


CHAPTER  II. 
FUNDAMENTAL    OPERATIONS. 

Addition. 

30.  Addition  is  the  operation  of  finding  the  simplest  equiva- 
lent expression  for  the  aggregate  of  two  or  more  algebraic 
quantities.     Such  expression  is  called  their  sum. 

31.  When  the  Terms  are  Similar  and  have  Like  Signs. 

+       a 

(1)  What  is  the  sum  of  a,  2a,  3a,  and  4a  /  4-    2a 

Take  the  sum  of  the  coefficients,  and  annex  the  common  -|-  3  a 
unit  or  literal  part.     The  first  term  (a)  has  a  coefficient  1      j_    4^ 

understood  (3  13).  

+  10a 

2ah 

(2)  What  is  the  sum  of  2a5,  ^ah,  6  ah,  and  ah  f  ^^^ 

Note.  —  When  no  sign  is  written,  the  sign  +  is  understood 

ab 


(?5). 


\2ab 


Add  the  following :  — 

(3)  a  (5)      lac  (7)  -2>ahc  (9)  ~2adf 

bac  —2abc  —Qadf 


a 


+  2a  12a6'  —bahc  —8adf 

(4)      8  ah  (6)    +^ahc       (8)   -3ac?  (10)  —    9abd 

7  ah  Sahc  —2ad  —I5ahd 

Ibah  +  7ahc  —  bad  —  24.ahd 

41 


42  ELEMENTARY    ALGEBRA. 

Hence,  when  the  terms  are  similar  and  have  like  signs, 
Add  the  coefficients,  and  to  their  sum  prefix  the  common 
sign.      To  this  annex  the  common  unit  or  literal  pari. 


Exercises. 

9  aZ>  +    ax 

2.    Sac' -3b' 

3.    Ibab'c'- 12 a.bc' 

Zah  +  Zax 

7ac'~8b' 

I2ab'c'-15abc' 

12ab  +  4:ax 

Sac' -9b' 

ab'c'  -      abc' 

32.   When  the  Terms  are  Similar  and  have  Unhke  Signs. 

The  signs  +  and  —  stand  in  direct  opposition  to  each  other. 

If  a  merchant  writes  +  before  his  gains,  and  —  before  his 
losses,  at  the  end  of  the  year  the  sum  of  the  pins  numbers  will 
denote  the  gains,  and  the  sum  of  the  minus  numbers  the 
losses.  If  the  gains  exceed  the  losses,  the  difference,  which  is 
called  the  algebraic  sum,  will  be  plus ;  but  if  the  losses  exceed 
the  gains,  the  algebraic  sum  will  be  minus. 

(1)  A  merchant  in  trade  gained  $1500  in  the  first  quarter 
of  the  year,  $3000  in  the  second  quarter,  but  lost  $3000  in  the 
third  quarter,  and  $800  in  the  fourth.  What  was  the  result 
of  the  year's  business  ? 


1st  quarter +  1500 

2d        "        +  3000 


+  4500 


3d  quarter -  3000 

4tli      "        -    800 


-3800 


+  4500  -  3800  =  +  700,  or  $700  gain. 

(2)  A  merchant  in  trade  gained  $1000  in  the  first  quarter, 
and  $2000  the  second  quarter.  In  the  third  quarter  he  lost 
$1500,  and  in  the  fourth  quarter  $1800.  What  was  the  result 
of  the  year's  business? 


Ist  quarter +  1000 

2d        ♦•       +  2000 


+  3000 


3d  quarter -  1500 

4th      "        -1800 


-3300 


+  3000  -  3300  =  -  300,  or  $300  loss. 


ADDITION.  43 


(3)  A  merchant  in  the  first  half  year  gained  a  dollars,  and 
lost  b  dollars.  In  the  second  half  year  he  lost  a  dollars,  and 
gained  b  dollars.     What  is  the  result  of  the  year's  business  ?' 


1st  half  year -\-  a 

2d         "  -a 

Result 0 


Ist  half  year —  h 

2d         "  +6 

Result 0 


Hence  the  algebraic  sum  of  a  positive  and  a  negative  quan- 
tify is  their  arithmetical  difference,  with  the  sign  of  the 
greater  prefixed. 

Add  the  following  :  — 
(4) 


Sab 

(5) 

4ac6^ 

(6)  -4a'&V 

Sab 

-  8acb' 

+  6aVc' 

-&ab 

acb' 

-2a'bV 

bab  —  3  acb"^  0 

Hence,  when  the  terms  are  similar  and  have  unlike  signs, 

Write  the  similar  terms  in  the  same  column. 

Add  the  coefficients  of  the  additive  terms  and  also  the  co- 
efficients of  the  sub  tractive  terms. 

Take  the  difference  of  these  sums,  prefix  the  sign  of  the 
greater  J  and  then  annex  the  literal  part  or  unit. 

Exercises. 

What  is  the  sum  of 

1.    2a'b'  -  ba'b'  +  7a'b'  +  6aV  -  lla'b'? 

2aV 

Having  written  the  similar  terms  in  the  same  column,  Oao 

we  find  the  sum  of  the  positive  coefficients  to  be  15,  and  4"     t  o,  b 

the  sum   of  the  negative  coefficients  to  be  — 16.     Their  -J-    6  a'^b^ 

difiference  is  —  1 :   hence  the  sum  is  —  a^b^.  11  a^b^ 

-    ^' 


44  ELEMENTAKY    ALGEBRA. 

2.  Sa'b  +  5a'b-3a'b  +  4:a'b-'6a''b  —  a''b^  Ayis.  2arb. 

3.  l2a!'bc'-^a'bc'  +  ^aJ'bc''-^a'bc'  +  lla'bc^'^   Ans.l7a'bc\ 

4.  4:a'b-8d'b-9a'b  +  lla'b?  Ans.~2a'b. 

5.  7abc^-abc^-7abc^  +  Sabc''  +  6abc'?  Ans,  I3abc\ 

6.  9cb^-5cb^  —  Sac'  +  20cb'  +  9ac^-24:cb^?    Ans.  +  ac\ 

33.   To  Add  any  Algebraic  Quantities. 

(1)  What  is  the  sum  of  3 a,  5 ^,  and  —2c? 

Write  the  quantities  thus:  3a +  56  — 2c,  which  indicates  their  sum, 
as  the  terms  are  dissimilar;  that  is,  have  no  common  unit. 

(2)  Let  it'  be  required  to  find  the  sum  of  the  quantities 

2a'-4:ab 
Sa'-Sabi-    b' 
2ab-5b' 


5a'~^ab-W 


From  the  preceding  examples  we  have,  for  the  addition  of 
algebraic  quantities,  the  following  rule  :  — 

Write  the  quantities  to  be  added,  placing  similar  terms  in 
the  same  column,  and  giving  to  each  its  proper  sign. 

Add  each  column  separately,  and  then  annex  the  dissimilar 
terms  with  their  proper  signs. 

Exercises. 
1.    Add  the  polynomials 
Sa''-2b'-4:ab,ba''-b'  +  2ab,  and  3aJ- 3c»-26*. 

The  term  3a^  being  similar  to  5a^  we  3L«  _  4LV,  _  0^2 

write  Sa^  for  the  result  of  the  reduction  of  p^i  »    1    ou?,  _     Ya 

these  two  terms,  at  the  same  time  slightly  '\    ~r  ^   Ya         ^ 

crossing  them,  as  in  the  first  term.  +^Vy  ~  2^^  —  3^ 

Passing  then  to  the  term  —  4a6,  which  Sa^  -\-     ab  —  5b^  —  36^ 
is  similar  to  -\-2ab  and  +3ab,  the  three 
reduce  to  +  a6,  which  is  placed  after  8  a*,  and  the  terms  crossed  like  the 


ADDITION.  45 

first  term.    Passing  then  to  the  terms  involving  6^,  we  find  their  sum  to 
be  —  5  6^,  after  which  we  write  —  3  c^. 

Note.  —  The  marks  are  drawn  across  the  terms,  that  none  of  them 
may  be  overlooked  or  omitted. 

2.         7ahc  +  9ax        3.      Sax  +  Sh  4.      12a--    6c 

—  Sahc  —  Sax  bax  —  9h  —    3a—    9c 


Aiabc  +  6ax  lZax  —  6h  9a— 15  c 

5.    If  a  =  5,  5  =  4,  c  =  2,  x  =  \,  what  are  the  numerical 
values  of  the  several  sums  above  found  ? 


6. 


9a+/           7. 
—  6a+g 
-2a- 

6ax—    8ac         8.    3q/*+     g+  m 

—  7  ax—    9ac                ag  —  Zaf—  m 

ax  ~{- 17  ac                ah—   ag-{-Sg 

lx  +  ?>ah  +  ^c 

—  ^x-Zah  —  bc 

bx  —  9ah  —  9c 

10.        8x'+    9acx+lSa'h'c^ 
-7x^-lSacx+Ua'h'c^ 
-Ax'+    4.acx- 20 a'h'c' 

11.        22A-3c?— 7/+3,^  12.        19  ah' +  Sa'h'-8ax' 

—  Sh  +  Sc-2f-9g  +  bx  -17 ah'- 9 a^h'  + 9 ax'' 

13.        7x-9g  +  bz  +  3-    g  14.        8a+    h 

—  x  —  Sy           —8—    g  2a—    5+    c 

—  x+    y  —  2>z+l  +  7g  -3a+    h          +2d 
-2x  +  6y  +  ?>z-l-    g  _65-3c  +  3cZ 

15.    Add   —h  +  ^c-d-llbe+6f-bg, 

U-2c-^d-e  +  27f,   bc-8d+Sf-7g, 
-7h-6c  +  17d  +  9e-bf+llg, 
^3b~bd-2e  +  Q-9g  +  h. 

Ans.  -8h-109e  +  S7f-10g  +  h. 


46  ELEMENTARY    ALGEBRA. 

16.  Add  the  polynomials  7a^b  —  S  abc  —  8  ^^c  —  9  c'  +  cd^, 

8abc  -  5a'b  +  3c'-  ib'c  +  cd\ 
and  4:a'b-Sc'  +  9b'c-M\ 

Ans.  6a'b  +  5abc-Sb'c-Uc'  +  2cd'~Sd', 

17.  What  is  the  sum  of  5  a'5c  +  6  ^o; -- 4  q/", 

-Sa'bc  —  ebx+Uaf,   - af+9bx  +  2a'bc, 

+  6af-Sbx  +  Qa'bc ?  Ans.  Y)a^bc -{-bx^  Ibaf. 

18.  What  is  the  sum  of  aW  +  3  a^m  +  b, 

-6aV  — 6a'm-6,  +  9b -9a'm- 5aW? 

Ans.  -  10  aV  -  12  a'm  +  9b. 

19.  What  is  the  sum  of  Aa^b^c  —  16a*x  —  9ax^d, 

+  6  a'b'c  -  6  ax'd  +  17a*x,  +  16  ax'd  -a*x-9  a'b'c  f 

Ans.  a^b'^c -{- ax^d. 

20.  What  is  the  sum  of  -1  g +  U +  4:g-2b +  2>g -^b 

+  2bf  Ans.  0. 

21.  What  is  the  sum  of  ab-{-^xy  —  m  —  n, 

-Qxy-^m+lln  +  cd,    +^xy  +  Am -lOn+fg  f 

Ans.  ab  +  cd  +fg. 

22.  What  is  the  sum  of  Axy  +  n  -\-  ^ ax  -\-  9 am, 

—  6xy  +  6n  —  6ax ~-  Sam^   2xy  —  ln  +  ax  —  amf 

Ans.  +  ax. 

23.  2{a  +  b)  24.  5(a^-c0  25.  9{c^~af) 
3(a  +  5)  _4(a^-c2)  7(c^-a/0 
2(a  +  ^>)                    -~l{a'-c')                    -  10 (c'  -  af) 

l{a  +  b)  6(c'-af) 

Note.  —  The  quantity  within  the  parentheses  must  be  taken  as  a 
whoU  (^  27).  In  Exercise  23  the  sum  of  a  and  b,  indicated  by  (a  +  h), 
is  the  unit;  in  Exercise  24  the  differenct  of  a*  and  c^  indicated  by 
(rt^  —  c'^),  is  the  unit. 


SUBTRACTION.  47 

26.  Add  3a(<7^  -  h')  -  2a{g'  -  h')  +  ^a{g'  -  h?) 

+  8 a (/  -h')-2a (g'  -  h').  Ans.  Ua(g'-  h'). 

27.  Add  3c(aV  -  ^'^)  -  9c(aV  -  ^2)  -  7c(a^(7  -  h') 

+  lbc {a^c  -h'')-\-c (a'c  -  b').  Ans.  3  c (a'c  -  b') . 

34.  In  algebra  the  term  "  add  "  does  not  always,  as  in 
arithmetic,  convey  the  idea  of  augmentation ;  nor  the  term 
"  sum  "  the  idea  of  a  number  numerically  greater  than  any  of 
the  numbers  added :  for,  if  to  a  we  add  —  b,  we  have  a  —  b, 
which  is,  arithmetically  speaking,  a  difference  between  the 
number  of  units  expressed  by  a  and  the  number  of  units 
expressed  by  b ;  consequently  this  result  is  numerically  less 
than  a.  To  distinguish  this  sum  from  an  arithmetical  sum, 
it  is  called  the  algebraic  sum. 

Subtraction. 

35.  Subtraction  is  the  operation  of  finding  the  difference 
between  two  algebraic  quantities. 

36.  The  subtrahend  is  the  quantity  to  be  subtracted ;  the 
minuend,  the  quantity  from  which  it  is  taken. 

37.  The  difference  of  two  quantities  is  such  a  quantity  as, 
added  to  the  subtrahend,  will  give  a  sum  equal  to  the  minuend. 

(1)  From  17  a  take  6  a. 

In  this  example,  17a  is  the  minuend,  and  Qa  the  subtra-  17  a 

hend.     The  difference  is  11a,  because  11a  added  to  6  a  gives  g^ 

17  a.  

1  -j 

Note.  ■ —  The  difference  may  be  expressed  by  writing  the 

quantities  thus  :  17a  —  6a=lla,  in  which  the  sign  of  the  subtrahend  is 

changed  from  4-  to  — . 

(2)  From  15x  take  —  9^:. 

The  difference,  or  remainder,  is  such  a  quantity  as,  being  15 X 

added  to  the  subtrahend  (—9  a:),  will  give  the  minuend  (15a;).  _    9^ 

That  quantity  is  24  a;,  and  may  be  found  by  simply  changing      

the  sign  of  the  subtrahend,  and  adding:    whence   we  may  24a; 
write,  15  a;  —  (—  9  x)  =  24  x. 


48  ELEMENTARY    ALGEBRA. 

(3)   From  10  ax  take  a  —  b. 

The  difference,  or  remainder,  is  such  a  quantity  as,  added  to  a  —  6, 
will  give  the  minuend  (10  ax).   What  is  that  quantity  ? 

If  you  change  the  signs  of  both  terms  of  the  subtra-     10  a^ 

hend,  and  add,  you  have  10ax  —  a-\-h.     Is  this  the  -j-a h 

true  remainder  ?     Certainly  :   for,  if  you  add  the  re-     — ' k 

mainder  to  the  subtrahend  (a  —  6),  you  obtain   the     lOax  —  a  +  bi 
minuend  (10  ax).  +  a  —  0 

It  is  plain,  that  if  you  change  the  signs  of  all  the     -ir) 
terms  of  the  subtrahend,  and  then  add  them  to  the 
minuend,  and  to  this  result  add  the  given  subtrahend,  the  last  sum  can 
be  no  other  than  the  given  minuend :  hence  the  first  result  is  the  true 
difference,  or  remainder  (§  37). 

From  the  preceding  examples  we  have,  for  the  subtraction 
of  algebraic  quantities,  the  following  rule  :  — 

Write  the  terms  of  the  subtrahend  under  those  of  the  minu- 
end, placing  similar  terms  in  the  same  column. 

Conceive  the  signs  of  all  the  terms  of  the  subtrahend  to  be 
changed  from  +  to  —  or  from  —  to  +,  and  then  proceed  as  in 
addition. 

Exercises. 

Subtract  the  following :  — 


1. 

Zab 

4. 

lea'b'c 

7. 

Sax 

2ab 
ab 

da'b'c 
Id'b'c 

Sc 

3ax-8c 

2. 

Q)ax 

5. 

17  a'b'c 

8. 

4iabx 

Zax 
?>ax 

Sa'b'c 
Ua'b'c 

9ac 

4  abx  —  9ac 

3. 

9abc 

6. 

l^aVx 

9. 

2  am 

7  abc 
2abc 

7a'b'x 
17  a' b\r 

ax 

2  am  —  ax 

SUBTRACTION.  49 

10.  From  9a'6'  take  3a'b\  Ans.  6d'b\ 

11.  From  IQa^xT/  take  —Iba^xy.  Arts.  Sld^xy. 

12.  From  12 ay  take  Say.  Ans.  4: ay. 

13.  From  19aVy  take  —  18aVy.  Ans.  37aVy. 

14.  From  Sa'b'  take  3a'6^  Ans.  Sa'b'-Sa'b''. 

15.  From  7a'^*  take  6a'b\  Ans.  7a'b'-6a*b\ 

16.  From  3ab^  take  a*'^>^  ^7i5.  3ab'-a''b\ 

17.  From  :r^y  take  y^x.  Ans.  a;'y  —  y'^a*. 

18.  From  3xy  take  a;y.  A7is.  Zxy  —  xy. 

19.  From  Sayx  take  a;yz.  ^ns.  Sd^y^x  —  xyz. 

20.  From  9a'^>'  take  —  3a'5^  ^ns.  I2a'b\ 

21.  From  14  ay  take  -  20 a'y.  ^ns.  34  ay. 

22.  From  -24a*^>^  take  l^a'b\  Ans.  -40a'b\ 

23.  From  —  13a:y  take  —  14:ry.  Ans.  xy. 

24.  From  —  47aVy  take  —ba^x^y.  Ans.  —  42aVy. 

25.  From  —  94aV  take  3aV.  Ans.  —  97aV. 

26.  From  a  +  x^  take  —y^.  Ans.  a-\-x^  +  y^. 

27.  From  a' +  ^)'  take  -  a' -  6^  ^ns.  2  a' +  25'. 

28.  From  —  16aVy  take  —  19aVy.  Ans.  +3aVy. 

29.  From  a^  ~  x"  take  d' +  x\  Ans.  —2  a:'. 

For  Review. 

Note.  —  Exercise  2  is  the  same  as  Exercise  1,  with  the  signs  of  the 
subtrahend  changed. 

Subtract  the  following :  — 

^1.         6ac  — 5a5+    c^  2.         ^ac  —  bab+    c^ 

wk  3ac  +  3a^>  +  7c  -3ac-3a^>-7c 


Zac-Sab-\-    c^~-1c  Zac-%ab+    c'-lc 

D.  K.  E.  A.  —  4. 


50                                      ELEMENTARY    ALGEBRA. 

3.        eax-a  +  Sb^ 
9ax-x+    h" 

5. 

-2a'  +  Sa'b-    Sb'c 

-2>ax-a^x-\-'ll;' 

7a'-la'b  +  Ub'c 

4.     ^yx  —  Zx'-rhh 
yx  —  S     +    a 

6. 

bab-~icd+3a'  +  5b' 

57/x~3x''  +  S  +  bb- 

—  a 

-ab  +  Scd-5b' 

1,    From  a  +  8  take  c~ 

5. 

Arts,  a  —  c  +  13. 

Arts.  —2>a}  —  ^b. 

9.  From  6 a;y  —  8 a?c^  take  —Ixy  —  d^c^.   Ans.  loxy—1  a^c^. 

10.  From  a  +  c  take  —a  —  c.  Ans.  2a  +  2c. 

11.  From  4  (a +  Z))  take  2(a  +  b).  Ans.  2(a  +  b). 

12.  From  3(a  +  :r)  take  (a  +  rr).  Ans.  2(a  +  x), 

13.  From  9 (a' -^r'^)  take  -2(d'-x').        Ans.  11(0" -x'), 

14.  From  6d'-lbb'  take  —3a'  +  9b\       Ans.  9a'-24Z>». 

15.  From  3a™-2Z>**  take  a"•-2^^^  Ans.  2dr, 

16.  From  9chn'  —  4:  take  4  — TcV/i'^  ^ns.  16cV-8. 

17.  From  6  am  +  y  take  3  a7?i  —  x.  Ans.  3  a??i  +  x-\-y. 

18.  From  3aa;  take  3ax  —  y.  Ans.  +y. 

19.  From-7/+3?7i-8a:  take  -  6/— 5m  -  2:r  +  3g?+ 8. 

Ans.  -f+Sm-6x-Sd-S. 

20.  From-a-5Z>  +  7c  +  cf  take  ib  -  c -{-2d+2Jc. 

Ans.  —a~9b  +  Sc-d-2k. 

21.  From  —  3a  +  ^>--8c+7e-5/+3A-7a:  — 13y   take 

^  +  2a  -  9c  +  8e  -  7a:  +  7/— 3/ -  3  Z  -  ;{:. 

A71S.  —5a  +  b  +  c-€~l2f+Sh-12y  +  3l. 

22.  From  2a;-4a-2^>  +  5  take  8-5Z)  +  a  +  6.r. 

Ajis.  — 4a:  — 5a  +  36  — 3. 


SUBTRACTION.  61 

23.  From  Sa  +  b  +  c  —  d—lO  td.'ke  c  +  2a  —  d. 

Ans.  a  -\-  b  —  10. 

24.  From  3a  +  ^>  +  c-c?— 10  take  ^>  — 19  +  3a. 

Ans.  c  —  d-{-d. 

25.  From  a^ -{-Sb^c  +  ab^  ~  abc  take  P  ~{- ab^  —  abc. 

Ans.  a^  +  Sb'c-b\ 

26.  From  12:r  +  6a-46  +  40  take  4Z>-3a  +  4a;+6^— 10. 

Ans.  Sx  +  9a  —  Sb  —  6d+50. 

27.  From2a;  — 3a  +  45  +  6c  — 50take9a  +  a;  +  6^>  — 6c— 40. 

Ans.  x-l2a-2b  +  l2c-'  10. 

28.  From  6a  — 4ib  —  12c +  12x  take  2x-8a  +  4:b -6c. 

Ans.  14a  — 85-6c+10a:. 

38,  In  algebra  the  term  "  difference  "  does  not  always,  as 
in  arithmetic,  denote  a  number  less  than  the  minuend :  for, 
if  from  a  we  subtract  —  b,  the  remainder  will  he  a  +  b;  and 
this  is  numerically  greater  than  a.  We  distinguish  between 
the  two  cases  by  calling  this  result  the  algebraic  difference. 

39.  When  a  polynomial  is  to  be  subtracted  from  an  algebraic 
expression,  we  inclose  it  in  a  parenthesis,  place  the  minus  sign 
before  it,  and  then  write  it  after  the  minuend.  Thus,  the 
expression  6a^  —  (Sab —  2b^ -{- 2bc)  indicates  that  the  poly- 
nomial 3 ab  —  2b^  -\-  2bc  is  to  be  taken  from  6a^.  Performing 
the  indicated  operations  by  the  rule  for  subtraction,  we  have 
the  equivalent  expression  6a^  —  Sab  ~\-2b^  —  2bc. 

The  last  expression  may  be  changed  to  the  former  by 
changing  the  signs  of  the  last  three  terms,  inclosing  them  in 
a  parenthesis,  and  prefixing  the  sign  — .     Thus, 

6d'-Sab  +  2b'-2bc  =  6a'-(Sab-2b'  +  2bc). 

In  like  manner  any  polynomial  may  be'  transformed,  as 
indicated  below. 


52  ELEMENTARY    ALGEBRA. 

7a'-Sa'b-4:b'c  +  6b'  =  7a'~-  (Sa'b  +  Wc-  6P) 
=  la'-Sa'b-(ib'c-6b'). 

Sa'  —  1  b'  +  c  -  d  =  8a'  -  0  b'  -  c  +  d) 

=  Sa'-7b'-(-c  +  d). 
9b^  -  a  +  S a"  -  d  =  9¥  -  (a  -  Sa'  +  d) 

=  9b'-a-{-Sa''  +  d). 
Note.  —  The  sign  of  every  term  is  changed  when  it  is  placed  within 
a  parenthesis  which  has  the  minus  sign  before  it,  and  also  when  it  is 
brought  out  of  such  parenthesis. 

40.  From  tlie  preceding  principles  we  have 

a  —  (+b)  —  a  —  b, 
and  a  —  (—b)  =  a  +  b. 

The  sign  immediately  preceding  b  is  called  the  sign  of  the 
quantity ;  the  sign  preceding  the  parenthesis  is  called  the  sign 
of  operation;  and  the  sign  resulting  from  the  combination  of 
the  signs  is  called  the  essential  sign. 

When  the  sign  of  operation  is  different  from  the  sign  of 
the  quantity,  the  essential  sign  will  be  — ;  when  the  sign  of 
operation  is  the  same  as  the  sign  of  the  quantity,  the  essential 
sign  will  be  +. 

Multiplication. 

41.  Multiplication  is  the  operation  of  finding  the  product  of 
two  quantities. 

The  multiplicand  is  the  quantity  to  be  multiplied  ;  the  mul- 
tiplier is  that  by  which  it  is  multiplied ;  and  the  product  is 
the  result.  The  multiplier  and  multiplicand  are  called  factors 
of  the  product. 

Exercises. 

1.  If  a  man  earns  a  dollars  in  1  day,  how  much  will  he 
earn  in  6  days  ? 

In  6  days  he  will  earn  six  times  as  much  as  in  1  day.  If  he  earns  a 
dollars  in  1  day,  in  6  days  he  will  earn  6  a  dollars. 


MULTIPLICATION.  53 

2.  If  1  hat  costs  d  dollars,  what  will  9  hats  cost  ? 

Ans,  9  d  dollars. 

3.  If  1  yard  of  cloth  costs  c  dollars,  what  will  10  yards 
cost?  Ans.  10c  dollars. 

4.  If  1  cravat  costs  h  cents,  what  will  40  cost? 

Ans,  405  cents. 

5.  If  1  pair  of  gloves  costs  b  cents,  what  will  a  pairs  cost? 

If  1  pair  of  gloves  costs  h  cents,  a  pairs  will  cost  as  many  times  h 
cents  as  there  are  units  in  a;  that  is,  b  taken  a  times,  or  ab,  which  de- 
notes the  product  of  6  by  a  or  of  a  by  b. 

6.  If  a  man's  income  is  3  a  dollars  a  week,  how  much  will 
he  receive  in  45  weeks? 

3ax46  =  12a6. 

If  we  suppose  a  =  4  dollars,  and  6  =  3  weeks,  the  product  will  be  144 
dollars. 

Note.  —  It  is  proved  in  arithmetic  (Davies'  *'  Standard  Arithmetic," 
?  50)  that  the  product  is  not  altered  by  changing  the  arrangement  of  the 
factors  ;  that  is,   \2ab  =  ax6xl2  =  6xaxl2  =  axI2x6. 

42.   To  find  the  Product  of  Two  Positive  Monomials. 

Multiply  3 a'5'  by  2o^h. 

We  write, 

3a262  X  2a'b  =  3x2x  a"  x  a^  xb'' xb  =  3x2aaaabbb; 

in  which  a  is  a  factor  4  times,  and  b  a  factor  3  times :  hence  (2  14) 

3  a'b^  x2a^b  =  3x2  a'b^  =  6  a^'b^, 

.  in  which  we  multiply  the  coefficients  together,  and  add  the  exponents  of 
the  like  letters. 

The  product  of  any  two  positive  monomials  may  be  found 
in  like  manner.     Hence  the  rule  :  — 

Multiply  the  coefficients  together,  for  a  new  coefficient. 

Write  after  this  coefficient  all  the  letters  in  both  monomials, 


54 


ELEMENTARY    ALGEBRA. 


giving  to  each  letter  an  exponent  equal  to  the  sum  of  its  ex- 
ponents in  the  two  factors. 

Exercises. 

1.  Sa'bc'x1ahd^r:=bQa^b''c'd\ 

2.  21  a^h'cd  X  8  ahc""  =  168  a'bVd, 

3.  4:abcx  7df=2Sabcdf 
Multiply  the  following :  — 


4.         Sa'b                6. 

6a:y2; 

8.        3a5V 

2a'b 

ayH 
6  arry^^;^ 

9a'Z)^c 

6a*b' 

27a^5V 

6.       Ud'x               7. 

a^xy 

9.       87aa;V 

I2x'y 

2xy^ 

3Z)Vy» 

144  aVy 

2a'xY 

261a^>V/ 

10. 

5a'b'x'  by  6cV. 

^725.  ZOa^b''&x^, 

11. 

10  a'b'c^  by  7acd. 

Ans.  TOa^Z^Vc^. 

12. 

36a'bVd'  by  20  abVd* 

^715.  120  a^b'c'd\ 

13. 

5a"»  by  3a^>^ 

Ans.  15a"+'^>\ 

14. 

3aH'  by  6a2^>^ 

^715.  18a"*+'^>"+'. 

15. 

Ga^A"  by  9a'b\ 

^715.  54a"»+^Z>"+^ 

16. 

ba'^b''  by  2a''Z>». 

^ws.  10a~+'^>"+«. 

17. 

Sa^'^V  by  2ab''c. 

^Tis.  lOa^+^^'^+V. 

18. 

Qa'b^'c''  by  3a'^>V. 

Ans.  18a^^>•"+V+^ 

19. 

20  a'b'cd  by  12aVy. 

^715.  2i0  a'b'cdx'y. 

20. 

14 a'^>«c^V  by  20aWy. 

Ans.  280  a' b'c'd*xy. 

21. 

8a^^V  by  7a*^T/. 

Ans.  b6a'b*xy\ 

MULTIPLICATION. 


55 


22.  I^axyz  by  b  cc'hcdx^y'^ . 

23.  64a*mVyz  by  8a&V. 

24.  9a^Z»Vc^'  by  12a'^V. 

25.  216a^V£/«by  3a'^>V. 

26.  lOa'b'c'dyx  by  Ua'b'c'dxy. 

43.   Multiplication  of  Polynomials. 

(1)  Multiply  a  —  bhyc. 

It  is  required  to  take  the  difference  between 
a  and  b,  c  times ;  or  to  take  c,  a  —  b  times. 

As  we  cannot  subtract  b  from  a,  we  begin  by 
taking  a,  c  times,  which  is  ac;  but  this  product 
is  too  large  by  b  taken  c  times,  which  is  be 
hence  the  true  product  is  ac  —  be. 

If  a,  b,  and  c  denote  numbers,  as  a  =  8,  &  =  3, 
and  c  =  7,  the  operation  may  be  written  in 
figures. 


Ans.  Slba^bcdx^y^z, 

Arts.  b\2a^b'^c^rrv'x*yz, 

Ans.  108a*^V(^l 

Ans.  MSa'b'^c^d^. 

Ans.  S^OaWc'dyxy. 


a—b 


■  b  hy  c  —  d. 

■  b  as  many  times  as 


(2)  Multiply  a 

It  is  required  to  take  a 
there  are  units  in  c  —  d. 

If  we  take  a  —  bc  times,  we  have  ac  —  be ; 
but  this  product  is  too  large  by  a  —  6  taken  d 
times,  a  —  b  taken  d  times  is  ad  —  db.  Sub- 
tracting this  product  from  the  preceding  by 
changing  the  signs  of  its  terms  (§  37,  rule),  we 
have  {a  —  b)-\-{a~-c)  =  ab  —  bc~ad-^  bd. 


'  —  be 


S  = 


56  -  21  =  35 


a~  b 

c  —  d 

ac  —  be 

—  ad-\-  bd 


ac  ~ 

be  —  ad-\-  bd 

8- 

-3        = 

5 

7- 

-2        = 

5 

56 

-21 
-16  +  6 

56  -  37  +  6  =  25 


From  the  preceding  examples  we  deduce  the  following 
rule :  — 

When  the  faetors  have  like  signs,  the  sign  of  their  product 
will  be  +. 

When  the  factors  have  unlike  signs,  the  sign  of  their  product 
will  be  —. 


56  ELEMENTARY    ALGEBRA. 

Therefore  we  say  in  algebraic  language  that  +  multiplied 
by  +,  or  —  multiplied  by  — ,  gives  + ;  —  multiplied  by  +,  or 
+  multiplied  by  — ,  gives  — . 

Hence  for  the  multiplication  of  polynomials  we  have  the 
following  rule :  — 

Multiply  every  term  of  the  multiplicand  hy  each  term,  of  the 
multiplier^  observing  that  like  signs  give  +,  ci'^d  unlike  signs 
— ;  then  reduce  the  result  to  its  simplest  form,. 

Exercises. 

Note.  —  All  the  terms  in  the  exercises  below  are  positive. 
Multiply  the  following :  — 
1.    3a'  +  4a^>  +  5^  by  2a  +  bh. 

3a^+   Aah  +h^ 

2a  +    bh 

6a^+    Sa'b+    2a})' 

6a^  +  23a'^  +  22a^>^  +  5^» 

44.  Note.  —  It  will  be  found  convenient  to  arrange  the  terms  of  the 
polynomials  with  reference  to  the  ascending  or  descending  powers  of 
some  letter :  that  is,  to  write  them  down  so  that  the  highest  or  lowest 
power  of  that  letter  shall  enter  the  first  term ;  the  next  highest  or 
lowest,  the  second  term  ;  and  so  on  to  the  last  term. 

The  letter  with  reference  to  which  the  arrangement  is  made  is  called 
the  leading  letter.  In  the  above  example  the  leading  letter  is  a.  The 
leading  letter  of  the  product  will  always  be  the  same  as  that  of  the 
factors. 

2.  x^  +  2ax  +  a'  by  a;  +  a.  Ans.  x^  +  3arr'  +  3a'a:  +  a\ 

3.  x"  +  y'  by  a;  +  y.  ^725.  a;*  +  ary*  +  x^y  +  y*. 

4.  3  aJ^  +  6  aV  by  3  aZ>^  +  3  aV^ 

Ans.  9  a^h'  +  27  a^^) V .+  18  aV*. 


MULTIPL1CATI0^^ 


57 


.  5.    a^h'  +  c^d  by  a  +  ^>.  Ans.  a^h^  +  ac'd  +  a'6»  +  he'd. 

6.    3  aa-'  +  9  a^)'  +  cd""  by  6  aV. 

'  Ans,  18a'cV  +  54aVZ>'  +  6aVJ\ 
.  7.    64aV  +  21a^x  +  ^ah  by  8aW. 

Ans.  bl2a^cdx^  +  216 a'ccZx  +  72 a*^>cc^. 

8.  a'  +  3a'a;  +  3aa;'^  +  .'r' by  a  +  .r. 

Ans.    a*  +  4  a^'a;  +  6  a V  +  4  a;r2  +  a:*. 

9.  a;^  +  y^  by  a;  +  3/.  Ans.    0(^  +  xy^ -\- x^y  +  y^ . 

10.  .r^  +  a;y®  -\- 1  ax  hj  ax  -{- b ax. 

Ans.  6  aa;«  +  6  aa:'3/«  +  42  aV. 

11.  a'  +  3a'i  +  3a5''  +  5'by  a  +  ^. 

Ans.  a*  +  4  a'5  +  6  a'^)*'  +  4  aZ)'  +  Z>\ 

12.  or*  +  ar^y  +  xy"^  +  3/^  by  a;  +  y. 

13.  ar'  +  2ar'  +  a;  +  3by  3a;  +  l. 

^715.  2>x'+1x^+bx^  +  \0x+^. 

For  Review. 
Multiply  the  following  :  — 
1.    2aa;-3a^>by  ?>x-h. 

2ax  —    Sab 
Sx    ~      b 


6  ax"^  —    9  abx 

-    2abx  +  Sab^ 

6ax^—ll  abx  +  3  ab^ 

2.  a*-2Phj  a-  b.  Ans.  a'  —  2ab'  -  a'b  +  2b\ 

3.  a;'  -  3a:  -  7  by  :r  -  2.  Ans.  x'  —  5x'-x+  14. 

A.   Sa*--bab  + 2b' hy  a' ~  7 ab. 

Ans.  Sa^~26a'b  +  S7a'b'^Uab\ 


58  ELEMENTARY    ALGEBRA. 

5.  j-2  J^h'  +  h^  by  h''  ~  1.  Arts,  b'  -  h\ 

6.  a;*-2rr''?/  +  4a;y  —  8a:y'  +  16y*  by  a;  +  2y. 

Arts.  r'  +  ?>2'i^, 

7.  4a;' -2?/  by  2y.  Am.  ^x'y  —  ^y' 

8.  2x  +  ^y  by  2.^  —  4?/.  ^ns.  4a;'  — 16y 

9.  x^  +  xhj  +  xy''  +  if  by  :r  —  y.  ^ns.  a;*  — y* 

10.  rr'^  +  a;?/  +  y'  by  :r'  —  xy  +  y^  ^ns.  rr*  +  xhf  +  y* 

11.  2a''-3a^  +  4^  by  5a'-6aa;-2:^;^ 

^Tis.  10  a*  -  27  a'o;  +  34  a V  -  18  aa;'  -  8a;* 

12.  3a;'-2a;y  +  5  by  a;'  +  2a;y-3. 

Ans.  3a;*  +  4ar'y  —  4a;'  -  4a;y  +  16  a;y  -  15. 

13.  3a;'  +  2a;y  +  3y'  by  2a;'-3a;y  +  5y'. 

Ans.  6a;''  —  5a;y  —  ^xS/  +  6a;y  +  15a;y  -  9a;y 

'       +10a;y  +  15y^. 

14.  8aa;  — 6  ai  —  c  by  2  aa;  + a^  +  c. 

Ans.  16  aV  -  4  a'hx  —  6  o?h''  +  6  aca;  -  7  ahc  -  c\ 

15.  3a'-5^>'  +  3c'  by  d'  —  h\ 

Ans.  3  a*  -  8  a'Z>'  +  3  aV  +  5  6*  -  3  ^V'. 

16.  3a'  — 55c^+    c/ 

-  ba^  +  Ud-d>cf 

-  15a*  +  ?>lo^hd-2^a^cf-  20 b'd'  +  Ubcdf-  Sc'f 

17.  arx  —  d}V'  by  aV.  ^ns.  a"'+ V+^  —  a*5 V. 

18.  dr-^-b""  by  a"*  —  ^'^  ^ns.  a""  —  ^>'r 

19.  dr-Y'b''  by  a"*  +  ^>r  ^n5.  a""  +  2 a*"5"  +  Z>^ 

Division. 

45,  Division  is  the  operation  of  finding  from  two  quantities 
a  third,  which,  being  multiplied  by  the  second,  will  produce 
the  first. 


DIVISION.  59 

The  first  is  called  the  dividend;  the  second,  the  divisor;  and 
the  third,  the  quotient. 

Division  is  the  converse  of  multiplication.  In  it  we  have 
given  the  product  and  one  factor  to  find  the  other.  The  rules 
for  division  are  just  the  converse  of  those  for  multiplication. 

46.   To  Divide  One  Monomial  by  Another. 

Divide  72  a' by  8  a^ 

72  a^ 
The  division  is  indicated  thus  : 


The  quotient  must  be  such  a  monomial  as,  being  multiplied  by  the 
divisor,  will  give  the  dividend:  hence  the  coefficient  of  the  quotient 
must  be  9,  and  the  literal  part  a^ ;  for  these  quantities  multiplied  by  8  a* 
will  give  72  a^.     Hence 

The  coefficient  9  is  obtained  by  dividing  72  by  8  ;  and  the  literal  part 
is  found  by  giving  to  a  an  exponent  equal  to  5  minus  3. 

For  dividing  one  monomial  by  another,  then,  we  deduce 
from  the  above  example  the  following  rule :  — 

Divide  the  coefficient  of  the  dividend  hy  the  coefficient  of  the 
divisor,  for  a  new  coefficient. 

After  this  coefficient  write  all  the  letters  of  the  dividend, 
giving  to  each  an  exponent  equal  to  the  excess  of  its  exponent  in 
the  dividend  over  that  in  the  divisor. 

47.  Signs  in  Division.  Since  the  quotient  multiplied  by  the 
divisor  must  produce  the  dividend,  and  since  the  product  of 
two  factors  having  the  same  sign  will  be  +,  and  the  product 
of  two  factors  having  different  signs  will  be  — ,  we  conclude, — 

(1)  When  the  signs  of  the  dividend  and  divisor  are  like,  the 
sign  of  the  quotient  will  be  +. 

(2)  When  the  signs  of  the  dividend  and  divisor  are  unlike, 
the  sign  of  4h^  quotient  will  he  —, 


60  ELEMENTARY    ALGEBRA. 

Again,  for  brevity  we  say, — 

+  divided  by  +,  and  —  divided  by  — ,  give  +. 
—  divided  by  +,  and  +  divided  by  — ,  give  — . 

±^  =  +  h;  =^=  +  «; 

+  a  —b 


+  a 


Exercises. 


+  9a'bc  +3abc 

Divide  the  following :  — 

5.  Ibax^y^  by  —3  ay.  Ans.  —hx^y^. 

6.  84a5':r  by  12  ^>'.  ^ns.  7aia;. 

7.  ~36a'Z)V  by  9a'Z)V.  ^ns.  -4a^>V 

8.  -99a*^)V  by  lla'Z^V.  Ans.  -9ab'x. 

9.  108a;y2'  by  54  0:^2;.  Ans.  2x2/^z\ 

10.  64a;y2*  by  —  16a:y2:^  ^tw.  —  4a:yz. 

11.  -96a.^^V  by  12a'^Z>c.  ^7^5.  —  8a'^>V. 

12.  -38a*ra*  by  2a'b'd.  Ans.  -Idabd". 

13.  -64a^Z>V  by  32a'^>c.  Ans.  -2ab'c\ 

14.  128 aVy^  by  I6axy\  Ans.  8aVy». 

15.  -256a*Z)V^^  by  16a'bc\  Ans.  -I6ab'c'd\ 

16.  200a®mV  by  —  50a'7?2?2.  ^47i5.  —  4amn. 

17.  300ar'yV  by  60xfz.  Ans.  hx^z. 

18.  27a^^>V  by  --9a6c.  Ans.  —Za^bc. 

19.  64a^?/V  by  32a?A'.  Ans.  2a^yz. 


DIVISION.  61 

20.  -SS  a^h'c^  by  11  aWc\  Arts.  -Sa'b'c^ 

21.  77aVV  by  -llaVV.  Ans.-7. 

22.  84:a*b'c'd  by  -  42  a*b'c'd.  Arts.  -2 

23.  -88a'5V  by  8a'^V.  ^ns.  -lla^> 

24.  16a;2  by  -^x.  Ans.  -2x 

25.  -88a"5'^  by  UaTb.  Ans.  -Sa^"*"^ 

26.  TYa'"^**  by  —  lla'^Z)'.  ^7?s.  -7a"*-'^"-' 

27.  84  a'^"*  by  42a'*^>^  Ans.  2a«-"^>"»-*. 

28.  -88a^Z>«  by  8a"Z>"*.  ^ns.  -lla^-'^Z^'" 

29.  96a5^  by  48a"5'^.  Ans.  2a^-'*^)^-^. 

30.  168 a; V  by  12a:"?/"*.  ^ns.  14a;"- V""* 

31.  256  a6V  by  16a"^>"c^  ^ns.  16a''-'*6^- V"^, 

48.  It  follows  from  the  preceding  rules  that  the  exact  divis- 
ion of  monomials  will  be  impossible 

(1)  When  the  coefficient  of  the  dividend  is  not  exactly/  divisi- 
ble by  that  of  the  divisor. 

(2)  When  the  exponent  of  the  same  letter  is  greater  in  the 
divisor  than  in  the  dividend. 

(3)  When  the  divisor  contains  one  or  more  letters  not  found 
in  the  dividend. 

In  any  of  the  above  cases  the  quotient  will  be  expressed  by 
a  fraction. 

A  fraction  is  said  to  be  in  its  simplest  form  when  the  numer- 
ator and  the  denominator  do  not  contain  a  common  factor: 

for  example,  12a^b'^cd  divided  by  Sa'^bc'^  gives    "^     ^ — -,  which 

Sa^bc^ 

may  be  reduced  by  dividing  the  numerator  and  denominator 

by  the  common  factors,  4,  a^  b,  and  c,  giving 

12a*b'cd_Sa'bd,     .      25a'b'd' _  5a 
Sa'bc'  2c    '  ^^^   Iba'b'd*     Sb'd 


62  ELEMENTARY    ALGEBRA. 

Hence,  for  the  reduction  of  a  monomial  fraction  to  its  sim- 
plest form,  we  have  the  following  rule  :  — 

Suppress  every  factor,  whether  numerical  or  literal,  that  is 
common  to  both  terms  of  the  fraction.  The  result  will  be  the 
reduced  fraction  sought. 

Exercises. 
48a^^W^      ^ad^  ^      1  a'b         1 


1. 


2. 


Q>a'bc'd'      Qa'd 
Divide 

5.  49a^^V  by  Ua'bc\ 

6.  ^amn  by  Zabc, 

7.  l^a^b^mn^  by  I2a'b'cd, 


Sdd'b'c'de      Sbce  Ua'b'     2ab 

Slab'c'd     S7  b'c  ,     4:a'b^     2  a 


6ab* 

3  b' 

Ans. 

7  be' 
2a 

Ans. 

2mn 
be 

Ans.  :^ 

wn' 

2aVcd 
8.    2Sa'bVd^  by  16ab'cdhn.  Ans,  ^~''''' 


9.    72a'c'b'  by  12 a'c^b'd.  Ans. 

10.    100 a^b^xmn  by  25a'b'd,  Ans, 


4:b'm 

6 
d'c'bd 
^a^bxm.n 
d 

11.  96a^5Vcyby  75aVary.  Ans.  ^-^'^'^'^/. 

2b  xy 

12.  Sbm'nfxY  by  15am*nf  Ans.  lll^^. 

Sam' 

13.  I21d'xy  by  I6d*xy.  A71S.      ^^^ 


i6dxy 


DIVISION.  63 

49.  In  dividing  monomials  it  often  happens  that  the  ex- 
ponents of  the  same  letter  in  the  dividend  and  divisor  are 
equal,  in  which  case  that  letter  may  not  appear  in  the 
quotient.  It  might,  however,  be  retained  by  giving  to  it  the 
exponent  0. 

If  we  have  expressions  of  the  form 

a     a^    a^     a*     a^       . 
~'    "o*    ~v    ~v    ~k''    ^tc, 
a    or    or    a*    or 

and  apply  the  rule  for  the  exponents,  we  shall  have 


"±  =  0}-'  =  a\  ^  -  a'-'  =  a\-^  =  a'-'  =  a',  etc. 
a  a  a"* 


«_^l-l 0     CL' 


But  since  any  quantity  divided  by  itself  is  equal  to  1,  it 
_  follows  that 

^  =  a'  =  l,  ^  =  a'-'  =  a'=l,  etc.; 
a  a^ 

or,  finally,  if  we  designate  the  exponent  by  m,  we  have 

a_  ^^m-m  ^  ^0  ^  ]L       ^^^^  ig 

a"* 

The  0  power  of  any  qua/ntity  is  equal  to  1 :  therefore 
Any  quantity  7)iay  he  retained  in  a   term,  or  introduced 
into  a  term,  by  giving  it  the  exponent  0. 


Exercises. 

1.  Divide  Ga'^SV  by  2aV. 

§i^  =  3  a'-V-'c'  =  3  a'b'c'  =  3  c\ 
2d^o^ 

2 .  Divide  8  a'b'c'  by  -  4  a^b'c.         Ans.  -  2  a^5 V  =  -  2  c\ 

3.  Divide  —32m^nVy'^  by  4  m Wo:?/. 


Ans.  —  Sm^n^xy  =  —  Sxy. 


64  KLEMENTARY    ALGEBRA. 

4.  Divide  -  96  a^b'c*"  by  -  24 a'b\  Ans.  4  a'5V**  =  4  c\ 

5.  Introduce  a  as  a  factor  into  6b^c*.  Ans.  6a%^c*, 

6.  Introduce  ab  as  factors  into  Qc^c?*.  Ans.  Qa^^Vc?**. 

7.  Introduce  abc  as  factors  into  Sd*/"*.  Ans.  Sa^b^c^d*/'*. 

60.   When  the  exponent  of  any  letter  is  greater   in   the  , 
divisor  than  it  is  in  the  dividend,  the  exponent  of  that  letter 
in  the  quotient  may  be  written  with  a  negative  sign.     Thus, 

?L'  =  i ;  also  ^'  =  a^-5  =  a-^  by  the  rule  : 


hence 


a' 
1 
a' 


Since  a  '  ~  — ,  we  have  5  X  a  '  =  —  ; 

that  is,  a  in  the  numerator  with  a  negative  exponent  is  equal 
to  a  in  the  denominator  with  an  equal  positive  exponent. 
Hence 

An2/  quantity  having  a  negative  exponent  is  equal  to  the  re- 
ciprocal of  the  savie  quantity  with  an  equal  positive  exponent. 

Any  factor  may  be  transferred  from  the  denominator  to  the 
numerator  of  a  fraction,  or  the  reverse,  by  changing  the  sign  of 
its  exponent. 


Exercises. 

1.   Divide  32a'5c  by 

16  a^^^ 

Ans. 

16a'6'      ^"    "    '^      a'b 

„     64a^5V      ^    „27_, 

A        6c 

3.   Keduce  IJ-^. 

4       x~h           z 
Ans.  -— — ,  or  -— -• 
3           Sx' 

DIVISION.  65 

4.    In  bay~^x~^  get  rid  of  the  negative  exponents. 

^*    -'^^       _3;_3  get  rid  of  the  negative  exponents. 

^^5,  —— -• 

6.    In         _3^_^  _^  get  rid  of  the  negative  exponents. 


3aV(^^ 


7.  Keduce Ans.  ,  or 

14a^6-^c  7         '  7a^ 

8.  Reduce  72a'^)'-f-8a^^>^  ^tw.  9a-'5-\  or  4- 

ah 

^'    -^^     g  get  rid  of  the  negative  exponents. 

ha-h-  3^, 

ac 

10.    Reduce  ^^^ — - — tt—^'  -4^5.  3a5V. 

—  ba-^h-^ 

51.   To  Divide  a  Polynomial  by  a  Monomial. 

Divide  each  term  of  the  dividend  separately  hy  the  divisor. 
The  algebraic  sum  of  the  quotients  will  be  the  quotient  sought. 

Exercises. 
Divide 

1.  ^a'b'^-a  by  a.  Arts,   ZaV—l. 

2.  ba'b'-2ba'b''  by  ha^b\  Ans.  l-ba, 

3.  ?>ba^b''  —  2bab  by  —bah,  Ans,  —lab  +  b. 
10  ah  — lb  ac  by  5  a.  Ans,  25  — 3  c. 

.    ^ah  —  ^ax  +  ^a^y  by  2a.  An^.  3J  — 4a;  +  2ay. 


n 


66  ELEMENTARY    ALGEBRA. 

6.  -lbax^  +  6x^  by  —Sx.  Arts,  bax-2x'. 

7.  -2l3:y'  +  35a'%-7cV  by  —  7y.     Am.  Sxi/-ba'b^+c\ 

8.  40 a'b'  +  8 a'b'  -  32 a*b'c'  by  8 a'b\     Am.  5  aM-  ^'  -  4 c\ 

52.   Division  of  Polynomials. 

(1)   Divide  -  2a +  6a'^- 8  by  2  + 2a. 

Dividend.         Divisor. 
6a'^-2a-8|  2a  +  2 
6a^  +  6a  3a  —  4  Quotient. 

-Sa-S 
-8a  — 8 


0  Remainder. 

We  first  arrange  the  dividend  and  divisor  with  reference  to  a  (§  44), 
])lacing  the  divisor  on  the  right  of  the  dividend.  Divide  the  first  term 
of  the  dividend  by  the  first  term  of  the  divisor.  The  result  will  be  the 
first  term  of  the  quotient,  which,  for  convenience,  we  place  under  the 
divisor.  The  product  of  the  divisor  by  this  term  {Qd^  -\-  6  a),  being  sub- 
tracted from  the  dividend,  leaves  a  new  dividend,  which  may  be  treated 
in  the  same  way  as  the  original  one ;  and  so  on  to  the  end  of  the 
operation. 

Since  all  similar  cases  may  be  treated  in  the  same  way,  we 
have,  for  the  division  of  polynomials,  the  following  rule  :  — 

Arrange  the  dividend  and  the  divisor  with  reference  to  the 
same  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  tei'm  of  the 
divisor,  for  the  first  term  of  the  quotient.  Multiply  the  divisor 
by  this  term  of  the  quotient,  and  subtract  the  product  from  the 
dividend. 

Divide  the  first  term  of  the  re^nainder  by  the  first  term  of  the 
divisor,  for  the  second  term,  of  the  quotient.  Multiply  the 
divisor  by  this  term,  and  subtract  the  product  from  the  first 
remainder,  and  so  on. 


DIVISION. 


67 


Continue  the  operation  until  a  remainder  is  found  equal  to  0, 
or  one  whose  first  term  is  not  divisible  hy  that  of  the  divisor. 

Notes. —  1.  When  a  remainder  is  found  equal  to  0,  the  division  is 
exact. 

2.  When  a  remainder  is  found  whose  first  term  is  not  divisible  by 
'  the  first  term  of  the  divisor,  the  exact  division  is  impossible.  In  that 
case,  write  the  last  remainder  after  the  quotient  found,  placing  the 
divisor  under  it,  in  the  form  of  a  fraction. 

(2)    Let  it  be  required  to  divide 

5la'b'  +  10a'-iSa'b-lbb*  +  4.ab'  by  ^ab-5a'  +  Sb\ 
Note.  —  First  arrange  the  dividend  and  divisor  with  reference  to  a. 
Dividend.  Divisor. 


10a*-48a'^  +  51a'^^»^+    4:ab'--ldb* 
10a*-    Sa'b-    6a'b' 


-ba'  +  4:ab  +  Sb' 


-  40  a'b  +  57  a'b'  +    4ab' -  156* 
-4:0a'b  +  32  a'b'  +  24  ab' 


■  2a' +  Sab -5b' 
Quotient. 


25  a'b'- 20  ab' - 
25a'b' -20  ab'- 


15  b' 
15  b' 


(3)     x'  +  x^y  +  x'y  +  xy''  - 

x'  +  x^y 

+  x'y  +  xy' 
+  x'y  +  xy"" 


■2y 


^  +  y 


a;'+  xy  - 


■2y 


Note.  —  Here  the  division  is  not  exact,  and  the  quotient  is  fractional. 


(4) 


1  + 
1- 


l  +  2a  +  2aM-2a'  +  ,  etc. 


+  2a 

'\-2a-2a' 
+  2d' 
-Y2a^- 


2a» 


+  2a« 

Note.  —  In  this  example  the  operation  does  not  terminate:  it  may  be 
continued  to  any  extent. 


68-  ELEMENTARY   ALGEBRA. 

Exercises. 
Divide 

1.  a'^  +  2ax  +  x^  by  a  +  x.  Ans.  a  +  x. 

2.  a'  —  3 a^y  +  3 ay^  —  y^  by  a  —  y.        Ayis.  a^  —  2ay  +  y^. 

3.  24:a'b  — 12 a'cb^- 6 ab  by  -6ab. 

Ans.  -4a  +  2aV5  +  l. 

4.  6x'-96  by  3a: -6.  ^ns.  2^;^  4a;'  + 8^;+ 16. 

6.    a^  — 5a*a:  +  10aV— 10aV+5aa;*— a;^  by  a'-2a2;  +  a:*. 

Ans.  a^  —  3  a^o:  +  3  ax^  —  x^. 

6.  48r'-76aa;'-64a'a;  +  105a'  by  2a: -3a. 

Ans.  24  a:'  — 2  ao:- 35  a'. 

7.  y*  —  3yV  +  3yV  —  a:«  by  y'  -  3y'a:  +  3ya;'  -  or*. 

Ans.  y'  +  3y'a:  +  3ya:'  +  a;*. 

8.  64  a*b'  -  25 a'b'  by  8 a'b'  +  5 a5^         ^ns.  Sa'b'  -  5 a5*. 

9.  6a^  +  23a'^6  +  22a5'  +  5^>'  by  Sa'  +  4:ab  +  b\ 

Ans.  2a  +  bb. 

10.  6aa:*  +  6aa:V  +  42aV  by  aa:  +  5aa;. 

Ans.  x^  +  xy^  +  7  ax. 

11.  -lba*  +  37a'bd-29a'cf-20b'd'  +  4:ibcdf-ScT  by 

3a'-5Z>G?+c/.  ^ns.  -  5a'  +  45c;- 8c/. 

12.  a:*  +  a:y  +  y*  by  x^  —  xy-\-y\  Ans.  a:'  +  a:y  +  y'. 

13.  a:*  —  y*  by  a;  —  y.  ^n6?.  oi?  +  a:'y  +  ry'  +  y*. 

14.  3a*  -  8a'6'  +  3aV  +  56*  -  3iV  by  a'  -  5'. 

^ns.  3a' -56'  + 3c'. 

15.  6a:«  -  5.ry  -  6a:y  +  6.ry  +  15a;y  -  9a:'y*  +  lO^iY 

+  15/  by  3a:H  2a:y  +  3y'. 

Ans.  2a:'-3a:'y'  +  5y\ 


DIVISION.  69 

16.  _  c'  +  16 aV  —  7  a6c  -  4  a^hx  -  6aW  +  ^acx  by   ^ax 

—  ^ah  —  c,  Ans.  2ax  +  ab  +  c. 

17.  Sx*  +  ^x^y  —  4lx^  ~  4a:y  +  163:3/  —  15  by  2xy  +  x^  —  S. 

Ans.  3x^  —  2x7/  +  5. 

18.  x^  +  32y^  by  x  +  2y, 

Ans,  x^-2x'y  +  4:xy-Sxy^+iey\ 

19.  Sa'-26a'b-Uab'  +  37a'h'  by  25=^- 5a^)  +  3al 

-4ns.  a^  —  7  ai. 

20.  a*  —  ^>*  by  a'  +  a'Z>  +  ctb'  +  Z^l  ^ns.  a  —  5. 

21.  o;^  —  3a;^y  +  y^  by  x  +  y.     Ans.  x^  —  4:xy  +  4:y^ ^. 

x  +  y 

22.  l  +  2a  by  I  — a -a'.    Ans.  1  +  3a  +  4a'  + 7a'+,  etc. 


CHAPTER  III. 

FACTORING,   GREATEST   COMMON    DIVISOR,    AND 
LEAST  COMMON   MULTIPLE. 

Useful  Formulas. 

53.  A  formula  is  an  algebraic  expression  of  a  general  rule 
or  principle. 

Formulas  serve  to  shorten  algebraic  operations,  and  are  also 
of  much  use  in  the  operation  of  factoring.  When  translated 
into  common  language,  they  give  rise  to  practical  rules. 

The  verification  of  the  following  formulas  affords  additional 
exercises  in  multiplication  and  division. 

54.  Formula  1.  —  To  form  the  square  of  a  +  Z>,  we  have 

(a  +  hy  =  {a  +  h){a  +b)  =  a^  +  2ah  +  h''  =  a^  +  h"  +  2ah ; 

that  is, 

The  square  of  the  sum  of  any  two  quantities  is  equal  to  the 
sum  of  their  squares,  plus  tivice  their  product. 

(1)  Find  the  square  of  2a  +  3^. 
We  have  from  the  rule, 

(2a  +  3  6)2  =  4a2  +  962-|-i2a6. 

(2)  Find  the  square  of  bab-\-  Sac. 

Ans,  25  a'b'  +  9  aV  +  30  a'bc, 

(3)  Find  the  square  of  5a^  +  Sa'^b. 

Ans.  2ba'  +  e4:a*b'  +  S0a'b. 

(4)  Find  the  square  of  6  ao:  +  9  a V. 

Ans.  36  aV  +  81  aV  +  108  aV. 
70 


FORMULAS.  71 

Note.  —  If  the  expression  to  be  squared  consists  of  more  than  two 
terms,  it  may  be  written  in  the  form  of  a  binomial,  and  squared.     Thus, 

(a  +  6  -f  c)2  =  [a  -f  (6  -f  c)f  =  a^  +  (6  +  cf  +  2a(6  +  c) 
=  a^  +  b^  +  c^  +  2ab  +  2ac  +  2bc; 
{a  +  b  +  c+dy  =  [{a  +  b)  +  {c  +  d)f 

=  (a  +  by  +  (c  +  df  +  2{a  +  b){c  +  d) 

=  a^  +  b^  +  c'^  +  d^  +  2ab  +  2ac  +  2ad  +  2bc-{-2bd  +  2cd. 

Find  the  square  of  the  following :  — 

(6)   2a  +  3b  +  c.       (6)  2x  +  37/  +  4:b.       (7)  x  +  a-\-2i/, 

65.   Formula  2.  —  To  form  the  square  of  a  difference,  a  —  b, 
we  have 

(a  -  by  =  (a  -  b)(a  -b)  =  a' -2ab  +  b' =  a'  +  b' -2ab  ; 

that  is, 

The  square  of  the  difference  of  any  two  quantities  is  equal  to 
the  sum  of  their  squares,  minus  twice  their  product 

(1)  Find  the  square  of  2a—  b. 

We  have  {2a-  bf  =  ^a"  +  6^  -  4a5. 

(2)  Find  the  square  of  4  ac  —  be. 

Ans.  16aV  +  ^>V-8a5c'. 

(3)  Find  the  square  of  7  a'b'- 12  ab\ 

Ans.   49  a'b*  +  144  a'b'  -  168  a'b\ 

Note.  —  Similarly, 

(a  -  6  +  c)2  =  [a  _  (6  -f  c)f  =  0^ +  {b -cf -2a{b -e) 

'^a^  +  b'^-\-(^-2ab-{-  2ac -  2bc; 
and 

(a  ~  6  -  c  -  cf)2  =  [(a  -  6)  -  (c  +  d)f 

=  (a  -  bf  i.(c~dy-2{a-  b){c  +  d) 

=  a^  -^  b^  +  <^  -^  d'^  -  2ab  -  2 ac  -  2ad  +  2bc  -\-  2bd+  2cd. 


72  ELEMENTARY   ALGEBRA. 

66.   Pormula  3.  —  To  multiply  a  +  h  by  a  —  h,vfQ  have 

{a  +  b)x{a-b)  =  a'-h''', 
that  is, 

The  sum  of  two  quantities,  multiplied  by  their  difference,  is 
\  equal  to  the  difference  of  their  squares. 

(1)  Multiply  2c +  ^>  by  2c -b.  Ans,  4:C^  —  b\ 

(2)  Multiply  9  ac  + 3  5c  by  9ac  —  Sbc. 

Ans,  81aV-9&V. 

(3)  Multiply  8  a' +7  a^^  by  Sa^-7ab\ 

Ans.  64a«-49a'5*. 

Note.  —  To  multiply  a  +  &  —  cby  a  —  h  +  c,  we  have 
(a  +  b  -  c)  X  {a  -  b  +  c)  =  [a  +  (b  -  c)]x  [a  ~  {b  -  c)]  =  a^  -  (b  -  c)'. 

Multiply  together  the  following :  — 

(4)  X  -j-y  —  4:  and  a;  —  y  +  4. 

(5)  2a-b-d  2ind  2a  +  b  +  d. 

57.  rormnla  4.  —  To  multiply  a^  +  ab  +  b^  hj  a  —  b,  we  have 

(a'  +  ab  +  b%a  -b)  =  a'-  b\ 

58.  Pormula  5.  —  To  multiply  a'^  —  ab  +  b^hja  +  b,  we  have 

(a'  -ab  +  b'')(a  +  b)  =  a''  +  b\ 

59.  Formula  6.  —  To   multiply  together  a  +  b,  a  —  b,  and 
a^  +  Z>^  we  have 

(a  +  b){a  -  b){a'  +  b'')  =  a' -  b\ 

60.  Since  every  product  is  divisible  by  any  of  its  factors, 
each  formula  establishes  the  principle  set  opposite  its  number. 

(1)   The  sum  of  the  squares  of  any  two  quantities,  plus  twice 
their  product,  is  divisible  by  their  sum. 


FACTORING.  73 

(2)  The  su7n  of  the  squares  of  any  two  quantities,  minus 
twice  their  product,  is  divisible  hy  the  difference  of  the  quantities. 

{?>)  The  difference  of  the  squares  of  any  two  quantities  is 
divisible  by  the  sum  of  the  quantities,  and  also  by  their  dif- 
ference. 

(4)  The  difference  of  the  cubes  of  any  two  quantities  is  divisi- 
ble by  the  difference  of  the  quantities  ;  also  by  the  sum  of  their 
squares,  plu^  their  product, 

(5)  The  sum  of  the  cubes  of  any  two  quantities  is  divisible 
by  the  sum  of  the  quantities ;  also  by  the  su7n  of  their  squares 
minus  their  product. 

(6)  The  difference  between  the  fourth  powers  of  any  two 
quantities  is  divisible  by  the  sum  of  the  quantities,  by  their 
difference,  by  the  sum  of  their  squares,  and  by  the  difference 
of  their  squares. 

Factoring. 

61.  Factoring  is  the  operation  of  resolving  a  quantity  into 
factors.  The  principles  employed  are  the  converse  of  those 
of  multiplication.  The  operations  of  factoring  are  performed 
by  inspection.     Thus, 

(1)   What  are  the  factors  of  the  polynomial  ac-\-  ab  -\-  ad  f 

We  see  by  inspection  that  a  is  a  common  factor  of  all  the  terms : 
hence  it  may  be  placed  without  a  parenthesis,  and  the  other  parts 
within.     Thus,   ac  ■\-  ah  -\-  ad  =  a{c  -\-  h  -\-  d). 

^     (2)   Find  the  factors  of  the  polynomial  a^b'^  +  a^d  —  af. 

Ans.  a'(b'  +  d-f), 

(3)  Find  the  factors  of  the  polynomial  Sa^b  —  6  aV  +  b^d. 

Ans.  biSa'-ea'b  +  bd). 

(4)  Find  the  factors  of  Ba'b  -da'c-  IS a'xy. 

A71S.  Sa^  (b  —  Sc  —  6  xy). 


74  ELEMENTARY    ALGEBRA. 

(5)  Find  the  factors  of  Sd'cx  -  18  acx'  +  2ac^7/  -  30  aV. 

Ans.  2ac{^ax  —  ^x'^  +  &y  —  15 aV). 

(6)  Factor  Z^a'h\-^o!'h''d}  ^-\8o^h''c\ 

(7)  Factor  12c?*W  -  15c«(^*- 6cW. 

(8)  Factor  lbo?bcf~lOahc^ -2bahcd. 

Ans,  bahc{%aj-2e^-bd). 

62,  When  two  terms  of  a  trinomial  are  squares,  and  posi- 
tive, and  the  third  term  is  equal  to  twice  the  product  of  their 
square  roots,  the  trinomial  may  be  resolved  into  factors  by 
Formula  1.     Thus, 

Factor  the  following :  — 

(1)  a^  +  2ah  +  h\  Ans,  (a  +  S)(a  +  J). 

(2)  ^a'  +  l2ah  +  9h\  Ans,  {2a  +  U){2a  +  U). 

(3)  9a^  +  12a5  +  4^>^  Ans,  (3a  +  26)(3a  +  2i). 

(4)  4a;^  +  8:r  +  4.  Ans,  (2a;  +  2)(2a;  +  2). 

(5)  ^a^b''  +  l2ahc  +  ^c\  Ans.  {^ah +  2c){?>ab  +  2c). 

(6)  l^xy  +  \^xf  +  ^y\  Ans,  {^xy +  2y''){^xy +  2y'). 
Note.  —  Factor  the  following  by  note  to  Formula  1. 

(7)  a^  +  2ab  +  2ac  +  b''  +  c'  +  2bc-=a^  +  b''  +  c''  +  2ab 

+  2ac  +  2bc.  Ans,  (a  +  b  +  c){a  +  b  + c). 

(8)  x^  +  ^a^  +  W  +  ^ax  +  ^bx+l2ab, 

Ans.  {x  +  2a  +  U){x  +  2a  +  2>b), 

63.  When  two  terms  of  a  trinomial  are  squares,  and  posi- 
tive, and  the  third  term  is  equal  to  minus  twice  their  square 
roots,  the  trinomial  may  be  factored  by  Formula  2.     Thus, 


FACTORING.  75 

Factor  the  following :  — 

(1)  o}  —  2ab  +  b\  Ans.  (a-b){a-b). 

(2)  4:a'-4:ab  +  b\  Ans.  (2a- b)(2a- b), 

(3)  9a'  +  c'-6ac.  Ans.  (3a  -  c)(3a- c). 

(4)  aV-4aar  +  4.  Ans.  {ax-2Xax-2). 

(5)  4:X^  +  y^  —  Axy.  Ans.  {2x  —  y)(2x —  y), 

(6)  36:r^-24rr3/  +  4yl  Ans.  (6x  -  2y)(6x -2y), 
Note.  —  Factor  the  following  by  note  to  Formula  2. 

(7)  a'-2ab  +  b'-2ac  +  c'  +  2bc  =  a'  +  b'  +  c'-  2ab 

—  2ac  +  2bc. 
Ans.  (a-b-  c)(a  -b  -  c)  =r.[a-{b  +  c)]  [a  -  (5  +  c)]. 

(8)  ^x'-^xyi-12bx  +  y^  +  9b^—6by. 

Ans.  (2x  —  y  +  Sb)(2x-y  +  3b) 

=  [2x-(y-3b)][2x-(y-Sb)l 

64.  When  the  two  terms  of  a  binomial  are  squares,  and  have 
contrary  signs,  the  binomial  may  be  factored  bv  Formula  3. 
Thus, 

Factor  the  following :  — 

(1)  4c'-b\  Ans.  (2c  +  b)(2c~b). 

(2)  81aV-9Z)V.  Ans.  (9ac  +  SbcX9ac--Sbc). 

(3)  64:a'b'-25xy.         Ans.  (Sa'b' +  6xyXSa'b' -  bxy). 

(4)  25a'c'-9xy,  An^.  (5ac  +  Bx'y)(5ac-Sx'y). 

(5)  d6a'b'c'-9x\  Ans.  (Qa'b'c  +  S:i^)i6a'b'c- 3x'). 

(6)  49a:*- 36y*.  Ans.  (1  x' +  67^X7 x"  -  6y').^ 
Note.  —  Factor  the  following  by  note  to  Formula  3. 

(7)  ^a'-ie(b-cy.    Ans.[2a-^(b-c)][2a  +  A(b-c)]. 

(8)  (x-Syy-16m\     ^ns.  (rr-3y+4?n)(.r-3y-4m). 


76  ELEMENTARY    ALGEBRA. 

65.  When  the  two  terms  of  a  binomial  are  cubes,  and  have 
contrary  signs,  the  binomial  may  be  factored  by  Formula  4. 
Thus, 

Factor  the  following  :  — 

(1)  Sa'-c\  Ans.  (2  a- c)(4a*  +  2ac  +  c'). 

(2)  27a«~64.  Ans.  (3  a- 4)(9a^  + 12a+ 16). 

(3)  a'-64:b\  Ans.  (a-4:bXa'  +  4:ab  +  l6b'). 

(4)  a'-27b\  Ans.  (a-SbXa" +  Sab +  9b'). 

66.  When  the  terms  of  a  binomial  are  cubes,  and  have  like 
signs,  the  binomial  may  be  factored  by  Formula  5.     Thus, 

Factor^  the  following :  — 

(1)  8a'  +  c\  Ans.  (2a  + c)(4:a' -2ac  + c'). 

(2)  27a^  +  64.     •  ^W5.  (3a+4)(9a^-12a  +  16). 

(3)  a'  +  6U\.  Ans.  (a  +  UXa' -4:ab +  16b'). 

(4)  a'  +  27b\  Ans.  (a  +  SbXa' -Sab +  9b'). 

67.  When  the  terms  of  a  binomial  are  4th  powers,  and  have 
contrary  signs,  the  binomial  may  be  factored  by  Formula  6. 
Thus, 

What  are  the  factors  of 

(1)  a'~b'?  Ans.  (a  +  bXa-bXa'  +  b'). 

(2)  Sla*-I6b*?       Ans.  (3a  +  25)(3a- 2i)(9a''  +  4i*). 

(3)  IQa'b'- 81  c'd'? 

Ans.  (2ab  +  S  cdX2  ab~S  cc^)(4  a'b^  +  9  c'cP). 

(4)  8aV-625cy? 

Ans.  \2ax  +  bcyX2ax  -  5cy)(4aV  +  2bcy). 


GREATEST    COMMON    DIVISOR.  77 

67a.    Verify  this  formula  by  multiplication  :  — 
{x  +  a){x  +  h)-=x''  +  {a  +  b)x  +  ab. 
By  it  the  following  expressions  may  be  factored :  — 

(1)  a;^+17:r+70.  Arts.  {x+l){x+lO). 

(2)  x""  -l^x  +  80.  Ans,  (x  -  S)(x  -  10). 

(3)  x'  +  bx-Se.  Ans,  (x  +  9)(x-4:). 

(4)  x'-Sx-3S.  Ans.  (x  +  3)(x~U). 

Greatest  Common  Divisor. 

68,  A  common  divisor  of  two  quantities  is  a  quantity  that 
will  divide  each  of  them  without  a  remainder.  Thus,  Sa^b 
is  a  common  divisor  of  9  a'^b'^c  and  3  a^b^  —  6  aV, 

69.  A  simple  or  prime  factor  is  one  that  cannot  be  resolved 
into  any  other  factors. 

Every  prime  factor  common  to  two  quantities  is  a  common 
divisor  of  those  quantities.  The  continued  product  of  any 
number  of  prime  factors  common  to  two  quantities  is  also 
common  divisor  of  those  quantities. 


a 


70.  The  greatest  common  divisor  of  two  quantities  is  the  con- 
tinued product  of  all  the  prime  factors  which  are  common  to 
both. 

71.  When  both  quantities  can  be  resolved  into  prime 
factors  by  the  method  of  factoring  already  given,  the  greatest 
common  divisor  may  be  found  by  the  following  rule :  — 

Mesolve  both  quantities  into  their  prime  factors. 
Find  the  continued  product  of  all  the  factors  which  are  com- 
mon to  both,  and  it  will  be  the  greatest  common  divisor  required. 


78  ELEMENTARY    ALGEBRA. 

Exercises. 

Note.  —  For  convenience,  G.  C.  D.  will  be  used  to  denote  greatest 
common  divisor. 

Find  the  G.  C.  D.  of 

1.  nha^h'c  and  25  a5o?. 
Factoring,  we  have 

Iba^h^c  =  3  X  5  X  oaabbc, 
25  abd  =5x5  abd. 
The  factors,  5,  5,  a,  and  b  are  common  :  hence 

5  X  5  X  a  X  &  =  25  a6  =  the  divisor  sought. 
Verification.        7oa^b^c  -j-  25  a6  =  Sabc, 
25  abd  -r-  2oab  =  d ; 
and,  since  the  quotients  have  no  common  factor,  they  cannot  be  further 
divided. 

2.  d^  —  2ab-{-h^  and  a^  —  b^.  Ans.  a  —  b. 

3.  a'  +  2ab  +  b'  and  a  +  b.  Ans.  a  +  b. 

4.  aV  —  4:ax-\-4:  and  ax  — 2.  Ans.  ax  — 2. 
6 .    Sa'b  -  9 a'c- 18 a^xy  and  b'^c  -  3 ^>c'  -  6 bcxy. 

Ans.  b  —  3  c  —  6  xy. 

6.  ^arc  —  \acx  and  Zc^g  —  Zagx.  Ans.  a(a  —  x),  or  a^  —  ax. 

7.  4c'-12cx  +  9x'  and  4^c^-9x\  Ans.  2c -3a:. 

8.  x^~y^  and  x^  —  y"^.  Ans.  x  —  y. 

9.  4:c'  +  4:bc  +  b^  and  4c'' -&l  ^n5.  2c +  b. 
10.    25  aV  -  9  xy  and  5  ac(f '  +  3  dVy\     Aris.  5  ac  +  3 xy. 

71a.  When  the  quantities  cannot  be  readily  factored, 
another  method  of  finding  the  G.  C.  D.  is  used,  —  one  of  suc- 
cessive division,  depending  on  the  following  principles :  — 

(1)  Any  COMMON  factor  of  two  quantities  is  a  factor  of  their 
O.  C.  D. :  hence,  in  finding  the  G.  C.  D.,  any  common  factor 
that  is  apparent  m.ay  be  suppressed^  and  set  aside  as  a  factor 
of  the  G.  CD. 


GREATEST    COMMON    DIVISOR.  70' 

(2)  No  factor  that  is  not  common  can  be  a  factor  of  the 
G.  C.  D. :  hence  any  factor  in  either  'polynomial  that  is  not 
common  may  be  suppressed  and  disregarded.  So,  also,  either 
polynomial  may  be  multiplied  by  any  factor  that  is  not  con- 
tained in  the  other, 

(3)  Any  quantity  that  will  exactly  divide  two  other  quanti- 
ties will  divide  the  difference  between  any  two  multiples  of  those 
quantities. 

Thus,  a  —  b  will  exactly  divide  the  two  quantities 

a'-2ab  +  b''  and  a^-b\ 

Multiply  each  of  these  quantities  by  some  number  (that  is, 
take  multiples  of  them),  —  the  first  by  5,  and  the  second  by  3, 
giving  ba^-lOab  +  bb''  and  ^a^-^b\  The  diiference 
between  these  multiples  is  2a^  —  10a5  +  85^,  and  this  is 
exactly  divisible  by  a  —  b. 

(4)  Any  quantity  that  will  exactly  divide  the  difference 
between  any  ^multiples  of  two  quantities  and  one  of  the  quan- 
tities will  exactly  divide  the  other  quantity. 

Thus,  take  the  two  quantities 

c'  +  cW  +  cd''  +  d^  and  c'  +  2cd+  d\ 

Take  a  multiple  of  each,  multiplying  the  first  by  6,  and  the 
second  by  'Ic,  giving 

6/  +  6cW+6cc^^+6(^-'^  and  2c^  + 4cW+ 2cc^'. 

The  difference  of  these  multiples,  4c^  +  ^c^d+^cd"-  +  6(/\ 
is  divisible  by  c-\-d,  and  so  is  the  quantity  e^  +  2  cc?  +  o?^ ; 
c-\rd  will  also  divide  the  other  quantity,  &  +  c^d-\-  cd'^  +  d^. 

To  show  how  these  principles  may  be  applied,  take  the  two 
polynomials 

18a;'-18a:V+6a.y_63^  and  \2x^ -\hxy-^%y\ 


80  ELEMENTARY   ALGEBRA. 

These  are  arranged  according  to  the  descending  powers  of  the  same 
letter,  x. 

The  factor  3  may  be  suppressed  in  each,  and,  as  it  is  common,  is  to  be 
set  aside  as  a  factor  of  the  G.  C.  D.  (Principle  1),  leaving 

63^  —  6x^y  +  2 xy'^  —  2y^  and  ix^  —  bxy  -\-  y^. 

Multiply  the  first  of  these  by  2  (Principle  2),  to  make  ita  first  term  * 
exactly  divisible  by  the  first  term  of  the  second  polynomial,  and  avoid 
fractional  coefficients,  and  divide  the  result  by  the  second  polynomial. 
In  performing  this  division,  multiply  the  first  remainder  by  4  to  avoid 
fractional  coefficients. 


12a^'-12x^y+    ^xy^-  4^ 
12x^-15x^y+    Zxy^ 


4iX^  —  5xy  +  y^ 


3a;  +  3y 


3  x'^y  +      xy^  —    4  y^ 

12x^y+    4:xy^-l6y^ 

'[2x^y-Wxy^  -{-    3?/^ 

ldxy'-19y^ 

19a;v^—  19  y^  is  a  difference  between  multiples  of 

12 a;^  —  12x^y  +  4 xy^  —  4^/^  and  4ix'^  ~  5xy  +  y^. 

Any   quantity   that  will   divide   these  latter   quantities  must    divide 
19  xy^  —  19 y^  (Principle  3),  and  any  quantity  that  will  divide 

19  xij^  —  19  y^  and  ix^  —  5xy  +  y^ 

must  also  divide  12ar^  -  12x^y  +  ixy'^  -  iy^.    What,  then,  is  the  G.C.D. 
of  19  xy^  —  19  y^  and  4  x^  —  5  xy  +  y^  ? 

19  y  may  be  suppressed  and  disregarded  (Principle  2). 


4a;2 
4x^ 

-bxy  -\-y^\x- 
-'4iXy          4  X 

-y 
-y 

-  xy  +  y^ 

-  xy  +  y^ 

a?  — y  is  the  G.  C.  D.  of  19xy^  -  19y'  and  4a^  —  5xy  +3/';  and  hence 
it  is  the  G.C.D.  of  1 2  r»  -  12x2?/ +  4x3/2  _  4^/3  ^nd  4  x^  -  5  xy  +  t/''. 
This  multiplied  by  3,  the  common  factor  set  aside  in  the  beginning,  giv- 
ing 3x  —  3y,  will  be  the  G.  C.  D.  of  the  polynomials  given. 


GREATEST    COMMON    DIVISOR.  81 

As  all  other  examples  of  the  same  kind  can  be  performed 
in  a  like  way,  we  have  the  following  rule :  — 

Arrancje  the  polynomials  xulth  reference  to  the  same  letter^ 
and  suppress  all  monomial  factors  in  either  polynom^ial.  If 
any  factor  so  suppressed  is  common  to  the  two  polynomials^ 
set  it  aside  as  a  factor  of  the  greatest  cominon  devisor. 

Multiply  the  polynomial  containing  the  highest  power  of  the 
leading  letter  thus  prepared  by  such  a  factor  as  will  make  its 
first  term  exactly  divisible  by  the  first  term  of  the  other.  Divide 
the  first  by  the  second,  and  continue  the  division  until  the  greatest 
exponent  of  the  leading  letter,  in  the  remainder,  is  at  least  one 
less  than  in  the  divisor.  If  the  highest  power  of  the  leading 
letter  in  both  polynomials  is  the  same,  either  polynomial  may 
be  used  as  the  divisor. 

Take  the  divisor  as  a  neiv  dividend  and  this  remainder 
as  a  divisor,  and  proceed  as  before;  and  continue  until  a 
remainder  0,  or  07ie  independent  of  the  leading  lettet\  is  found. 
In  the  first  case,  the  last  divisor,  viultiplied  by  com7non  factors 
set  aside,  will  be  the  greatest  common  divisor ;  in  the  second^ 
the  product  of  factors  set  aside,  if  any,  luill  be  the  greatest  com- 
mon divisor.  If  there  be  no  remainder  0,  and  no  factors  to 
set  aside,  the  polynomials  are  prime  with  respect  to  each  other ^ 
and  have,  of  course,  no  comynon  divisor. 

(1)   Find  the  G.  C.  D.  of 

■B        3a;'-13a;^  +  23^-21  and  I2x^ +2x' -^^x  +  ^2. 

Suppressing  the  factor  2  in  the  second  polynomial,  and  introducing 
the  factor  2  into  the  first,  we  have 

6ifc3_26a;2  +  46a;-42  |6a.-^  +  a;^  -  44a;  +  21 
6.'c3+       a;2-44a;-f2l    1 

-27a;2  +  90a;~63 
r.  >'.E.  A.  —  6. 


82  ELEMENTARY   ALGEBRA. 

Suppressing  the  factor  —  9  in  the  first  remainder,  and  proceeding  as 
before,  we  have 

63^+      x'^-Ux-\-21  13a;'' -10a;  4- 7 
6r^-20x^  +  Ux  2a; +  7 

21a;2-58a;  +  21 
21a;2-70a;  +  49 


12a; -28 

Suppressing  the  fac1t>r  4  in  the  second  remainder,  and  proceeding  as 

before,  we  have 

3a-2-10a;-f-7  |3a;-7 
3a;2-    7a;  a;-l 


-  3a; +  7 

-  3a;  +  7 


0 
Hence  3  a;  —  7  is  the  G.  C.  D.  sought. 

(2)  x'  -  7x^  +  Sx'  +  28a:  -  48  and  x^  -  8x'  +  19:r  -  14. 

Ans.  X  —  2. 

(3)  4lx'  +  9x'  +  2x^-2x-^  and  Sa^  +  5x'-x  +  2, 

A71S.  x  +  2. 

(4)  x'  +  Sx*  -8x'-9x-S  and  x'  -  2x'  -  6x^  +  Ax" 

+  13x  +  Q,  Ans.  (x+lXx  +  lXx+l). 

(5)  6x^-4:X*-Ux'-dx'-Sx-l  and  ix'  +  2x'-lSx'' 

+  3a7  — 5.  A71S.  2x'  —  ^x^  +  x-l. 

(6)  20x''-l2af'  +  l6x'-15x'+Ux'-l5x  +  4: 

and  15a;*-9a:'  +  47a;^-21a;  +  28. 

Ans.  5x^  —  Sx  -\-4:. 

(7)  x^-x*-~Sx'—2x''+Ux  +  21,  x*  +  x^-2x  +  6. 

A71S.  x'  +  2x  +  3. 

To  find  the  G.  C.  D.  of  three  or  more  quantities, 

Find  the  G.  C.  B.  of  the  first  and  second,  then  the  O.  0.  D, 
oj  this  result  and  the  third  quantity,  and  so  on  to  the  last. 


LEAST    COMMON    MULTIPLE.  83 

Find  the  G.  C.  D.  of  the  following  :  — 

(8)  Aax^y,  l^ahx^,  and  2^acx^, 

(9)  ^x'-^x,  2a^-4:x\  and  x'i/-2x7/. 

(10)  x'-9x^  +  26x-2i,  :r^-10a;^  +  31a;-30, 

and  x'  -  llo;'^  +  38a;  -  40.  Ans.  x-2. 

(11)  x^-lOx'  +  d,  a:*  +  10a;^  +  20.r'^-10a;-21, 

and  x'  +  4:x'-  22^;'^  -4:X  +  2h  Ans.  x"  —  1. 

(12)  x*  —  a\  x^  +  a^,  and  x^  —  d\  Ans.  x  +  a. 

(13)  a'  +  b',  a'-h\  and  a'  +  b\  Ans.  a  +  h. 

(14)  '^x^—lx^y  +  bxAf~y\  x^y  ■\-Zxy" -Zx"" -f, 

and  Zt? -^hx^y-\-xy'^ —  y^.  Ans.  3x  —  y. 

Least  Common  Multiple. 

72.  A  multiple  of  a  number  is  any  quantity  that  can  be 
divided  by  that  number  without  a  remainder.  Thus,  Sa^b  is 
a  multiple  of  8,  also  of  d^  and  of  b. 

73.  A  common  multiple  of  two  or  more  quantities  is  a  quan- 
tity that  can  be  divided  by  each  separately  without  a  remainder. 
Thus,  24  aV  is  a  common  multiple  of  6  ax  and  4ia'^x. 

74.  The  least  common  multiple  of  two  or  more  quantities  is 
the  simplest  quantity  that  can  be  divided  by  each  without  a 
remainder.     Thus,  12a^Z)V  is  the  least  common  multiple  of 

^2a%  4a^)^and  6db'x\ 

75.  Since  the  common  multiple  is  a  dividend  of  each  of 
the  quantities,  and  since  the  division  is  exact,  the  common 
multiple  must  contain  every  prime  factor  in  all  the  quantities ; 
and,  if  the  same  factor  enters  more  than  once,  it  must  enter 
an  equal  number  of  times  into  the  common  multiple. 


84  ELEMENTARY    ALGEBRA. 

When  the  given  quantities  can  be  factored  by  any  of  the 
methods  already  given,  the  least  common  multiple  may  be 
found  by  the  following  rule  :  — 

Resolve  each  of  the  quantities  into  its  prime  factors. 

Take  each  factor  as  many  times  as  it  enters  any  one  of  the 
quantities,  and  form  the  continued  product  of  these  factors. 
It  will  he  the  least  common  multiple. 

Exercises. 

Note.  —  For  convenience,  L.  C.  M.  will  be  used  to  denote  least  com- 
mon multiple. 

1.  Find  the  L.  C.  M.  of  12a^5V  and  ^a'b\ 

12  a^b'c'  =  2. 2. 3.  aaabbcc. 
8a''b'  =  2.2.2.aabbb. 

Now,  since  2  enters  three  times  as  a  factor,  it  must  enter  three  times 
in  the  common  multiple ;  3  must  enter  once ;  a,  three  times ;  b,  three 
times ;  and  c,  twice :  hence  2.2.2.3  aaabbbcc,  or  24  a^6V,  is  the  L.  C.  M. 

Find  the  L.  C.  M.  of 

2.  6  a,  5a''b,  and  25abc\  Ans.  150 a''bc\ 

3.  Sa'b,  9abc,  and  27  aV.  Ans.  27 a'bcx\ 

4.  4aVy^  Sa^xy,  16  ay,  and  24aya;.  Ans.  48  a V/. 

5.  ax  —  bx,  ay  — by,  and  x^y^. 

Ans.  (a  —  b)x .x.yy  =  ax^y^  —  bx'^y^. 

6.  a  +  b,  a^-b\  and  a^  +  2ab  +  b\    Ans.  {a  +  b)\a-b). 

7.  3a''^>^  9aV,  ISaV,  ^3ay.  Ans.  l^a'h'xy. 

8.  8a\a~-b),  lba\a-b)\  and  12aV-^'). 

A71S.  \20a\a-b)\a-\-b). 


LEAST    COMMON    MULTIPLE.  85 

75  a.  When  the  given  quantities  cannot  readily  be  factored, 
another  method  is  used. 

The  L.  C.  M.  of  two  polynomials  contains  all  the  factors  of 
each  polynomial ;  the  G.  C.  D.  contains  all  the  factors  common 
.  to  the  two  polynomials.  Hence,  to  find  the  L.  C.  M.  of  two 
polynomials,  we  have  the  following  rule :  — 

Find  the  O.  C.  D.  of  the  two  quantities.  Divide  one  of  the 
quantities  hy  it^  then  TnuUiply  the  other  quantity  hrj  the 
quotient. 

Exercises. 

Find  the  L.  C.  M.  of  the  following :  — 

1.  2a7'^~a;y-6y'  and  Sa;'^- 80:3/  + 4y^ 

Their  G.  C.  D.  is  a;  —  23/ .-  hence  their  L.  C.  M.  is 

^x'-xy-^f  ^  ^3^2  _  g^^  ^  ^^2)  _  (2a;  ^  Zy){Sx^  -Sxy  +  4i/2). 
x  —  2i/ 

2.  Sx'-5x  +  2  Si,nd  4:ar'-4:x^~x+l. 

Ans.  (Sx  —  2)(4:a^  -  Ax'  -  X  +1). 

3.  6x'-x-l  and  2x^  +  Sx  —  2. 

Ans.  (^x+lX2x'  +  Sx-2), 

To  find  the  L.  C.  M.  of  several  quantities, 

Find  the  L.  C.  M.  of  the  first  and  second^  then  of  that  result 
and  the  third,  and  so  on  to  the  last. 

Find  the  L.  C.  M.  of  the  following :  — 

4.  x^-^x'  +  llx-^,  :x^-^x''  +  2^x-2A, 

and  a;' -  8a;'' +  19a; -12. 

Ans.  {x-Y){x-2){x-'Z){x-A). 

5.  a;*_10a;^  +  9,  a:*+10a;^  +  20a;^-10a;-21, 

and  a;*  +  4a;' -22a;' --4a; +  21. 

Ans.  (a:'-l)(ar'-9)(a:+7). 


86  ELEMENTARY    ALGEBRA. 

6.  x^  —  (a  +  h)x  +  ab,  x^  —  (b  +  c)x  +  hc^ 

and  x^  —  {c  +  a)x  +  ca.       Arts,  (x  —  a){x  —  h)(x  —  c). 

7.  e(a'-PXa-h)\  9(a'-b*Xa-b)\  and  12(a'-by. 

Arts.  36  (a'  -  b'){a'  -  b')(a'  -  bj, 

8.  x^  +  5x  +  6,  :r'-2a:-8,  and  rr'  — a:-12. 

Ans.  (x+S){x  —  4:Xx  +  2). 

9.  6:r'  +  13a;  +  6,  6a;'  — 5a;- 6,  and  Ax' -9. 

Ans,  (3a;  +  2)(2a;  +  3)(2a;-3). 

10.    2a;'  +  lla;  +  15,  2a;*  +  a:-10,  and  a;'  +  a;-6. 

Ans.  (2a;  +  5)(a;  +  3)(a;-2). 


CHAPTER  IV. 
FRACTIONS. 

76.  A  fractional  unit  is  any  one  of  a  number  of  equal  parts 

of  a  unit.     Thus,    -»    ->    -.    -)    are  fractional  units. 
2    4     7    5 

77.  A    fraction    is    a    fractional   unit,   or   a   collection    of 

fractional  units.     Thus,   ->    ->    -»    -,,    are  fractions. 
2     4:     7     b 

78.  Every  fraction  is  composed  of  two  parts,  —  the  denom- 
inator and  the  numerator.  The  denominator  shows  into  how 
many  equal  parts  the  unit  1  is  divided ;  and  the  numerator, 
how  many  of  these  parts  are  taken.     Thus,  in  the  fraction 

-,  the  denominator  b  shows  that  1  is  divided  into  b  equal 

parts,  and  the  numerator  a  shows  that  a  of  these  parts  are 
taken.  The  fractional  unit,  in  all  cases,  is  equal  to  the  re- 
ciprocal of  the  denominator. 

79.  An  entire  quantity  is  one  which  contains  no  fractional 
part.     Thus,  7,  11,  a^x,  4a;'^  — 3y,  are  entire  quantities. 

An  entire  quantity  may  be  regarded  as  a  fraction  whose 

denominator  is  1.     Thus,  7  =^-,  ab  =  — 


80.    A  mixed  quantity  is  a  quantity  containing  both  entire 

[id   fractic 
quantities. 


and   fractional   parts.      Thus,    7/^,   8f,   aH ,    are   mixed 


?? 


88  ELEMENTARY    ALGEBRA. 

81.  Let  -  denote  any  fraction,  and  q  any  quantity  what- 
ever.     From  the  preceding  definitions,  -  denotes  that  -  is 

0  0 

taken  a  times ;    also   -/■   denotes  that  -  is  taken  aq  times, 
0  b 

that  is, 

h      h     ^ 
Hence 

Multiplying  the  nuraerator  of  a  fraction  hy  any  quantity  is 
equivalent  to  multiplying  the  fraction  by  that  quantity. 

We  see,  also,  that 

Any  quantity  may  be  multiplied  by  a  fraction  by  multi- 
plying it  by  the  numerator,  and  then  dividing  the  result  bv  the 
denominator. 

82.  It  is  a  principle  of  division  that  the  same  result  will 
be  obtained  if  we  divide  the  quantity  a  by  the  product  of 
two  factors,  p  X  q,  as  would  be  obtained  by  dividing  it  first 
by  one  of  the  factors,  p,  and  then  -dividing  that  result  by  the 
other  factor,  q ;  that  is, 

^  _  (^  ^  _  AA 

Hence 

Multiplying  the  deno7ninator  of  a  fraction  by  any  quantity 
is  equivalent  to  dividing  the  fraction  by  that  quantity. 

83.  Since  the  operations  of  multiplication  and  division  are 
the  converse  of  each  other,  it  follows,  from  the  preceding 
principles,  that 

Dividing  the  nume7'ator  of  a  fraction  by  any  quantity  is 
equivalent  to  dividing  the  fraction  by  that  quantity. 

Dividing  the  denominator  of  a  fraction  by  any  quantity  is 
equivalent  to  multiplying  the  fraction  by  that  quantity. 


FRACTIONS.  89 

84.  Since  a  quantity  may  be  multiplied  and  the  result 
divided  by  the  same  quantity  without  altering  the  value, 
it  follows  that 

Both  terms  of  a  fraction  may  he  multiplied  by  any  quantity, 
or  both  divided  by  any  quantity,  without  changing  the  value 
of  the  fraction. 

Transformation  of  Fractions. 

86.  The  transformation  of  a  quantity  is  the  operation  of 
changing  its  form  without  altering  its  value.  The  term 
**  reduce  "  has  a  technical  signification,  and  means  "  to  trans- 
form." 

86.  To  reduce  an  Entire  Quantity  to  a  Fractional  Form  having 
a  Given  Denominator. 

Let  a  be  the  quantity,  and  b  the  given  denominator.     We 

have,  evidently,  a  =  —•     Hence  the  rule  :  — 

Multiply  the  quantity  by  the  given  denominator,  and  write 
the  product  over  this  given  denominator. 

87.  To  reduce  a  Fraction  to  its  Lowest  Terms. 

A  fraction  is  in  its  lowest  terms  when  the  numerator  and 
the  denominator  contain  no  common  factors. 

It  has  been  shown  that  both  terms  of  a  fraction  may  be 
divided  by  the  same  quantity  without  altering  its  value : 
therefore,  if  they  have  any  common  factors,  we  may  strike 
them  out.     Hence  the  rule  :  — 

Resolve  each  term  of  the  fraction  into  its  prime  factors,  then 
strike  out  all  that  are  common  to  both. 

The  same  result  is  attained  by  dividing  "both  terms  of  the 
fraction  by  any  quantity  that  will  divide  them  without  a 
remainder,  or  by  dividing  them  by  their  G.  C.  D. 


90 


ELEMENTARY    ALGEBRA. 


Exercises. 


Keduce  to  their  lowest  terms 
.     16  aV 


2bacd                      ^ 

Factoring,                              j^  r,   .     y 
2d  acd      5 .  5  acd 

Canceling  the  common  factors,  5,  a,  and  c,  we  have 

15  aV      3ac 

25  ace?      5  c? 

,      85Z)W^ 

Arts, 

mc'dT 

Ans. 

4. 


6. 


10. 


11. 


12. 


\2e'd^f 
ah  —  ac 
b-c  ' 
n^-2n  +  l 

x^  —  2ax-{-a^ 

96a'b'c 
- 12  a'b'c 

24b'-S6ab' 

a'-h' 
d'-2ab  +  b' 
ba^-lOa'b  +  bab' 

Sa'-Sa'b 
Sa'  +  ea'b' 


12a*  +  6aV 
a^  +  2ax  +  x'^ 


Ans. 


21 

5c 

dr 


Ans.  -  =  a. 


Ans. 

Ans. 
8 


n+1 
x" 


Ans. 


1 

U~Q>a 
^a'-lla'b'^' 


Ans. 


a±h 
a  —  b 

Ans.  ^J^^. 
8a 

1  +  25' 


Ans. 


\a^-\-2ae 


Ans. 


a^-  X 
Z(a-x) 


FRACTIONS.  91 

88.   To  reduce  a  Praction  to  a  Mixed  Quantity. 

"When  any  term  of  the  numerator  is  divisible  by  any  term 
of  the  denominator,  the  transformation  can  be  effected  by 
division.     Hence  the  rule  :  — 

PeiforTYi  the  indicated  div9sion^  continuing  the  operation  as 
far  as  possible ;  then  write  the  remainder  over  the  denominator^ 
and  anyiex  the  result  to  the  quotient  found. 

Exercises. 
Keduce 

Ans.  a 


1, 

ax  —  a^ 

X 

2. 

ax  —  x^ 

X 

3. 

ah--2a^ 
b 

4. 

a^-x" 

a~x 

5. 

x^-y" 

x-y 

,6. 

lOx^-bx  +  Z 

bx 

7. 

36.^-72a:  +  32aV 

^x 

8. 

l^acf-^bdcf-2ad 

3adf 

9. 

x'  +  X-4: 

x  +  2 

10. 

a'  +  b' 

a  +  b 

11. 

x^  +  Sx-26 

X 

Ans.  a  —  X, 

Ans.  a r— 

b 

Ans.  a-tx. 

Ans.  x^  -\-  xy-\-  y^. 

Ans.  2x-l+—' 
ox 

Ans.^x^-^  +  ^^^. 


.        6o     2bc      2 
Ans.  -— 


Ans.  x  —  1 
Ans.  a  —  b 


d        a       Sf 

2 


x  +  2 
2  b' 

'a  +  b 


Ans.  x+7  +  -^- 
.r  —  4  *  x  —  4: 


92  ELEMENTARY    ALGEBRA. 

89.   To  reduce  a  Mixed  Quantity  to  a  Fractional  Form. 
This  transformation  is  the  converse  of  the  preceding,  and 
may  be  effected  by  the  following  rule  :  — 

3Iultiply  the  entire  part  hy  the  denominator  of  the  fraction, 
and  add  to  the  product  the  m^ierator.  Write  the  result  over 
the  denominator  of  the  fraction. 

Exercises. 

Eeduce  the  following  to  fractional  forms :  — 

1.    6|. 

6x7f=42;  42  +  1  =  43:  hence  61  =  —. 

7 

XX  X 

3.    :.-^^  +  <  Ans.  ^^-^ 


2a  2 


a 


.     ^  ,  2.2:- 7  .        Vlx-1 

4.    5  +  —- Ans.  —- 

3a;  Zx 

a  a 

6.1  +  2:.-^.  Ans.  10^  +  4^  +  3. 

007  5a; 

7.    2a  +  J  -  3£±4.  ^,,.  16«  +  8S-3c-4 

8  8 

4a  4a 

9     8  I  3a;-     8  +  6a'y:r«  ,       96 a&x^  +  30 a'6V  -  8 

12a6a;«  12a6:i;* 


FRACTIONS.  93 

90.  To  reduce  Fractions  having  Different  Denominators  to  Equiv- 
alent Tractions  having  the  Least  Common  Denominator. 

This  transformation  is  effected  by  finding  the  L.  C.  M.  of 
the  denominators. 

13  5. 

Eeduce  -,  -,  and  —  to  their  least  common  denominator. 
3  4  12 

The  L.  C.  M.  of  the  denominators  is  12,  which  is  also  the  least  common 
denominator  of  the  required  fractions.  If  each  fraction  be  multiplied  by 
12,  and  the  result  divided  by  12,  the  values  of  the  fractions  will  not  be 
changed. 

-  X  12  =  4,  1st  new  numerator. 

o 

3 

-  X  12  =  9,  2d  new  numerator. 
4 

-1-  X  12  =  5,  3d  new  numerator. 
12 

4      9  5 

Hence  — ,    — ,    and  —  are  the  new  equivalent  fractions. 

12    12  12  ^ 

From  the  preceding  example  we  deduce  the  following 
rule :  — 

Mnd  the  least  common  multiple  of  the  denominators. 
Multiply  each  fraction  by  it,  and  cancel  the  denominator. 
Write  each  product  over   the   comnrion  multiple,   and  the 
results  will  he  the  required  fractions. 

Or,  in  general, 

Multiply  each  numerator  by  all  the  denominators  except  its 
[    own,  for  the  new  numerators ;  and  all  the  denominators  togethe7\ 
for  a  common  denominator. 


I 

94  ELEMENTARY   ALGEBRA. 

Exercises. 
Reduce  the  following  to  their  least  common  denominators ;  — 
1.   _,J^_  and      ' 


d'-b''  a  +  b 

The  L.  C.  M.  of  the  denominators  is  (a  +  h){a  —  h). 
■^^^  X  (a  +  h){a  -b)  =  a. 

— -  X  (a  +  h)(a  -h)  =  c(a-  h). 

Hence  ;; — and \  ~    ^  ,^   are  the  required  fractions. 

(a  +  h){a  -b)  (a  +  h){a  -h)  ^ 

„     3:r     4  ,    l^x" 

2.  — -)    -,    and 

4      6  15 

3.  a, ,    and 

4  6 

.     ^x     2b         .    , 
^'   TT-'    ^r-»    and  a. 
2a     3c 

^n.  "5.^-^r,    i^;il^,    and 


60 

:    40 
'    60' 

48  a:^ 
60 

^W6 

12a 
•12' 

95^ 
12' 

10  c^ 
12 

-4715. 

9c.r 
Qac 

4a5 
6ac 

6  ace? 
6ac 

^n5. 

9a 
12a' 

12  a 

12a;  +  24  a: 
12  a 

a;^ 

^      c      c  —  b  -, 

7.    -— ,    ,    and 


{l-xf       (l-xf  (l-xf 


5a        c  a-\-b 

A  ^ .       <^^^  +  i^c^  ba^c  —  b  a^b  +  5  abc  —  bab^           5  ac* 

Ans.  ■ >    ■ , 

ba^c  -\-  babe  bd^c  +  babe                 ba^e  +  babe 

Q        ex         dx^  J        ^ 

8. ,    ,  and 


a  —  xa-\-x  a-\-x 

Ans.  ^-?-(^+^),    M^Zli?:},    and  ^!X^lZ1^. 
a^  —  a;'  a*^  —  x^  d^  —  .t' 


FRACTIONS.  95 


Addition  of  Fractions. 

91.  Fractions  can  be  added  only  when  they  have  a  common 
unit ;  that  is,  when  they  have  a  common  denominator.  In 
that  case,  the  sum  of  the  numerators  will  indicate  how  many 
times  that  unit  is  taken  in  the  entire  collection.  Hence  the 
rule : — 

Reduce  the  fractions  to  be  added,  to  a  common  denominator. 
Add  the  numerators  together,  for  a  new  numerator,  and 
write  the  sum  over  the  coynmon  denominator. 

Exercises. 

1.  Add  -,    -,    and  -. 

By  reducing  to  a  common  denominator,  we  have 
6  X  3  X  5  =  90,  1st  numerator. 
4  X  2  X  5  =  40,  2d  numerator. 
2  X  3  X  2  =  12,  3d  numerator. 
2  X  3  X  5  =  30,  the  denominator. 
Hence  the  expression  for  the  sum  of  the  fractions  becomes 
90     40     12  _  142 
30     30     30      30* 
which,  being  reduced  to  the  simplest  form,  gives  4}^. 

2.  Find  the  sum  of  ^,    -,    and  -• 

V    d  f 

Here  ax  dxf  ^  adf  ^ 

c  xb  xf  =  cbf  > ,  the  new  numerators, 
exb  xd=ebd  ) 

i^and  b  X  dxf  =  bdf  the  common  denominator. 

iB    Hence  ff  +  '^-  +  f£=  ^d/+ch/+ebd  ^ 

bdf     hdf     bdf  bdf 


96  ELEMENTARY   ALGEBRA. 

Add  the  following  :  — 

^  Sx"^        ^    7    ,  2ax  .  ,   ,   ,  2abx  —  ^coi^ 

3.  a  —  --—   ^^^^   ^  H ^^5.  a  +  b-i —^^' 

0  c  be 

4.  -,    ^,    and   -•  ..^'— Am.  x  +  ^ 

2     3  4  '^  ^12 

5.  -^    and   -.  Ans.  — ^^-. 

6.  0;-+^   and    3x  +  ^^,  Ans.  4 :.  +  I^^-Hll. 

7.  4a:,   —— ,    and    — -^ —  Ans.  4:X-\ ■ ■ 


9 


a 


2x 


8.  ^,    I^,  and   2^+1.  ^...  2.  +  ^-^--±i^. 

3       4'  5  60 

9.  4:x,   ^,  and    2  +  f.  ^72s.  2  +  4a:  +  — • 

9.5  45 

10.  Sx  +  ?^  and    o;-^.  ^n5.  3a;  +  — • 

5  9  ^45 

11.  ac  — ■— -  and    1 — -•                 Ans.  1  +  ac —^ 

8a  a  8aa 

-o          3  3                ,4  .        4a:'^-5a;  +  4 

12.    , ,    and Ans.  ■ — 

(x~-if  (x-iy       x-i  i^-iy 

iQ      1  1.1  .1 

13.  — — -,     -— -,    and -•  Ans. 


4(1  + a)'    4(1 -a)'  2(1 -a^)  '  l-a' 

Subtraction  of  Fractions. 

92.  Fractions  can  be  subtracted  only  when  they  have  the 
same  unit ;  that  is,  a  common  denominator.  In  that  case,  the 
numerator  of  the  minuend,  minus  that  of  the  subtrahend,  will 
indicate  the  number  of  times  that  the  common  unit  is  to  be 
taken  in  the  difference.     Hence  the  rule  :  — 


FRACTIONS.  97 

Reduce  the  two  fractions  to  a  common  denominator. 

Then  subtract  the  numerator  of  the  subtrahend  from  that  of 

the  minuend,  for  a  new  numerator,  and  write  the  remainder 

over  the  com?non  denominator. 

Exercises. 

3  2 

1.    What  is  the  difference  between  -  and  -? 

7  8 

3_2_24_14_10_^ 

7      8      56      56      56      28* 

2-.    Find  the  difference  of  the  fractions  ^^  and  ^^~^^. 

2b  3c 

Here  {x  —  a)xSc  =  Scx  —  Sac^ 

,       ,    ,  >,  the  numerators ; 

{2a-4.x)x2b  =  4:ab-Sbxi 

and  26  X  3 c  =  66c,  the  common  denominator. 

Sex  — Sac     4:ab  —  Sbx     Sex —  Sac —  4:ab  +  Sbx 


Hence 


6bc  6  be  6bc 


3.  Required  the  difference  of  -^  and  ~  Ans.  — -- 

4.  Required  the  difference  of  5y  and  — ^-  Ans.  — r^- 

8  8 

5.  Required  the  difference  of  —  and  — •  Ans.  -— — 

6.  From  ^  +  ^  subtract  ^~^.  Ans.  4^' 

x  —  y  x  +  y  ^     y 

7.  From  subtract  — -•  Ans.  '^  \,^~    • 

y-z  y'^-z^  f-  z\ 

Find  the  differences  of  the  following  :  — 

o     ^x-^-a       .   2x-\-r  .        24^  + 8a -10^^-355 

8.  — — ^ —  and  — ——  Ans.  ' -7-7 

hb  8  406 

^     ci      ^  X       1  X  —  a  A        ct     ,  ex  -{-  bx  —  ab 

9.  Zx  +  -  and  x -^.  Ans.  2x-\ '—- 

b  c  be 

in     ^1      a  —  x  1      a-\-x  A^^    „         4:X 

10,    a-] — ; and  — -r-^ -•  Ans.  a — • 

a{a-\-  X)  a  {a  —  x)  a  —xr 

D.  N.  E.  A.  —  7. 


98  ELEMENTARY    ALGEBRA. 


Multiplication  of  Fractions. 

93.    Let  -  and  -  represent  any  two  fractions.    It  has  been 
0  a 

shown  (§  81)  that  any  quantity  may  be  multiplied  by  a  frac- 
tion by  first  multiplying  by  the  numerator,  and  then  dividing 
the  result  by  the  denominator. 

To  multiply  y  by  -,  we  first  multiply  by  c,  giving^;  then 
b         a  0 

we  divide  this  result  by  d,  which  is  done  by  multiplying  the 

denominator  by  d.     This  gives  for  the  product,  — -;  that  is, 

od 

a      c a£ 

b     d~bd 
Hence  the  rule  :  — 

If  there  are  mixed  quantities,  reduce  them  to  a  frojctional 
form;  then 

Multiply  the  numerators  together,  for  a  new  numerator ;  and 
the  denominators,  for  a  neio  denominator. 

Exercises. 

1.  Multiply  a  +  ^  by   -• 

a  d 

First,  a  +  ^  =  ^i^^: 

a  a 

hence  ^itjjg  x  -  =  ^^!i±^. 

a  d  ad 

Find  the  products  of  the  following  quantities :  — 

^     2.r     Sab  ^  Sac  •     .  An 

2.  — ,    ,    and  — — -  Ans.  9  ax. 

a        c  2b 

3.  5+^  and  2.  Ans.  2^±-^. 

a  X'  X 

.     x'-b'       .  x'  +  b'  '    ,  ,         x*-b* 

4.  — ; —  and  —-f —  Ans. 


be  b  +  c  '  b^c  +  bc^ 


FRACTIONS.  99 

,  x+1        J    X  —  1                     A        ax^  —  ax -\- x^  —  \ 
5.    x-\ ■ —   and   -•  Ans. 


a 


a  +  b  a^  +  ab 


^         ,     ax  J    a^  ~  x'^  ^        a'  +  a'^x 

6.  a-] and   -•  Aiis.  — ■ — — • 

a  —  X  x  +  x^  x-\-  x^ 

Multiply 

^       2a     .      d'-b' 

7.    _  by  — 

a  —  b     ^        3 

We  have,  by  the  rule, 

2a    ^a'-h'^  _2a{a^-h'^)  _2a{a-\-h)(a-h)  _2a.       ^. 
a-b  3  3(a-6)  S{a-b)  3   ^         ^* 

Note.  —  After  indicating  the  operation,  we  factored  both  numerator 
and  denominator,  and  then  canceled  the  common  factors,  before  perform- 
ing the  multiplication.  This  should  be  done  whenever  there  are  common 
factors. 

8.  _2_  by  ^l^Z^.  Ans.  2l£±i^). 
x  —  y             a  a 

9.  t^  bv  -i^.  An..  M^:^. 

3         -   ar  +  2  3 

10.  {2l±K  by  -14-  Ans.  2«(a+6). 

11.  (^^ll)!  by  (^+%'.  ^...  i^ll^. 

2/  «-!  2/ 

12.  (5^!zi£!l  by  1±^.  Anz.  ^^. 

1  —  a;^  a  +  a;  1  —  a; 

13.  a;  +  — ^  by  a; ~  Ans.  x\ 

^x-y     ^  x  +  y 

-.     2a  — b  1      6a  —  2b  .        ^  — 3a 

14.  — by  -- — --— -  Ans. 


4a        '  b^-2ab  '     2ab 

X     y  y,       x*~i 

^JL±.  Ans.  — — ^ 

y  ^  X  xhj 


15.    ^-|by^  + 


100  ELEMENTARY    ALGEBRA. 


Division  of  Fractions. 

V  1 

94,   Since  -  =/>  X  -,  it  follows  that  dividing  by  a  quantity 

is  equivalent  to  multiplying  by  its  reciprocal.     But  the  re- 
ciprocal of  a  fraction,  -,  is  -  (§  28) :  consequently,  to  divide 
Oj  c 

any  quantity  by  a  fraction,  we  invert  the  terms  of  the  divisor, 
and  multiply  by  the  resulting  fraction.     Hence 

b      d     b     c      be 

Whence  the  following  rule  for  dividing  one  fraction  by 
another :  — 

Reduce  mixed  quantities  to  fractional  forms. 

Invert  the  terms  of  the  divisor^  and  multiply  the  dividend  by 
the  resulting  fraction. 

Note.  —  The  same  Remarks  as  were  made  on  factoring  and  reducing, 
under  the  head  of  "  Multiplication,"  are  applicable  in  division. 

Exercises. 
Divide 

1.    a  —  —-  by  •^. 

h      2ac  —  b 
a = " 

2c  2c 

Hence  a- A  ^X  ^  ^oc^  ^  g^  2acy-- &y. 

2c     g         2c        f  2cf 


■1.. 


a 

-  "J 

i    - 

a 

2(a;  + 
a 

jH^ 

^ 

a 

/2{x  + 
a 

J^x 

i.x  + 

a 

-y) 

2 

a-y 

1x 
5 

by 

12 
13' 

Am. 

91a; 
60 

4«' 
7 

■by 

bx. 

Ans. 

4a; 
'  35* 

FRACTIONS.  101 

5.  ^+i  by  ^.  Ans.  ^+i. 

6.  -^  by  ^.  ^n5.  -^. 
a;-l     -^   2  a:-l 

7.  -X-   by  — -•  Ans.  — — 

^     X  —  b  -,      Sex  M        X  —  b 

8. by  — — •  Ans.  -— — -• 

Scd      ^    4:d  6c'x 

^     4.r^-8a;  .      a:^-4  .           4x 

9.   by  — - —  Ans.  -• 

8           ^       8  x+2 

10.  — —: —  by  — ' — — •  Ans.  x-\ 

x^  —  2bx  +  b'^x  —  b  X 

11.  2x(a  +  b)  by  -^.         .  Ans.  (^±il'. 

^          ^     ^  a  +  b  2x 

12.  i^!^  by  ^-  +  y  ^..  (^:=i)'. 

y                X  - 1  / 

13.  flzi^  by  1^^.  Ans.  i^Mnf!). 

14.  ^^^  by  i+^'.  ^ns.  ^^^^. 
1  — a;a  +  a;  1  —  x^ 

15.  :,^by  a:-^^.  ^^,.  ^±^. 
-^     5  — 3ai      6a  — 25  .        2a  — 5 

17.  Elr:/by^  +  ^.  ^n..  ^^'. 

x^y            y      X  X 

18.  m'  +  l+-^  by  m  +  i  +  1.  Ans.  m  +  --l. 

V    ^l  +  xyj         \           l  +  xyj  ^ 

20.    r^^±M  +  A  by  /^^+2^__^Y  ^^,.  1. 

\x  +  y     y)       \    y        x  +  yj 


CHAPTER  V. 

EQUATIONS   OF  THE   FIRST   DEGREE,    AND 
INEQUALITIES. 

Equations  of  the  Fikst  Degree. 

95.  An  equation  is  an  expression  of  equality  between  two 
quantities.  Thus,  x  =  b  +  c  is  an  equation  expressing  the 
fact  that  the  quantity  x  is  equal  to  the  sum  of  the  quantities 
b  and  c. 

96.  Every  equation  is  composed  of  two  parts,  connected  by 
the  sign  of  equality.  These  parts  are  called  members.  The 
part  on  the  left  of  the  sign  of  equality  is  called  the  first 
member ;  that  on  the  right,  the  second  member.  Thus,  in  the  • 
equation  x  +  a  =  b  —  c,  x  +  aia  the  first  member ;  and  b  —  c, 
the  second  member. 

97.  An  equation  of  the  first  degree  is  one  which  involves 
only  the  first  power  of  the  unknown  quantity.     Thus, 

6a:  +  3a;-5-13  (1) 

ax  -{- bx  -{-  c  =  d  (2) 

are  equations  of  the  first  degree. 

98.  A  numerical  equation  is  one  in  which  the  coeflicients  of 
the  unknown  quantity  are  denoted  by  numbers. 

99.  A  literal  equation  is  one  in  which  the  coefficients  of  the 
unknown  quantity  are  denoted  by  letters. 

Equation  (1)  is  a  numerical  equation ;   Equation  (2)  is  a 
literal  equation. 
102 


EQUATIONS    OF    THE    FIRST    DEGREE.  103 

Transfor7}iation  of  Equations. 

100.  The  transformation  of  an  equation  is  the  operation  of 
changing  its  form  without  destroying  the  equality  of  its 
members. 

101.  An  axiom  is  a  self-evident  proposition. 

102.  The  transformation  of  equations  depends  upon  the 
following  axioms :  — 

Axiom  1.  If  equal  quantities  be  added  to  both  members  of 
an  equation,  the  equality  will  not  be  destroyed. 

Axiom  2.  If  equal  quantities  be  subtracted  from  both  meyn- 
bers  of  an  equation,  the  equality  will  not  be  destroyed. 

Axiom  3.  If  both  members  of  an  equation  be  m^ultiplied  by 
the  same  quantity,  the  equality  will  not  be  destroyed. 

Axiom  4.  If  both  membeo^s  of  an  equation  be  divided  by  the 
same  quantity,  the  equality  will  not  be  destroyed. 

Axiom  5.  Like  powers  of  the  two  members  of  an  equation 
are  equal. 

Axiom  6.  Like  roots  of  the  two  members  of  an  equation  are 
equal. 

103.  Two  principal  transformations  are  employed  in  the 
solution  of  equations  of  the  first  degree,  —  clearing  of  fractions, 
and  transposing. 

Take  the  equation 

The  L.  C.  M.  of  the  denominators  is  12.  If  we  multiply 
both  members  of  the  equation  by  12,  each  term  will  reduce  to 
an  entire  form,  giving 

8a: -9a: +  22;  =  132. 


104  ELEMENTARY    ALGEBRA. 

Any  equation  may  be  reduced  to  entire  terms  in  the  same 
manner. 

104.  Hence,  for  clearing  of  fractions,  we  have  the  following 
rule :  — 

Find  the  L.  C.  M.  of  the  denominators. 
Multiply  both  memhers  of  the  equation  hy  it,  reducing  the 
fractional  to  entire  terms. 

Notes.  —  1.  The  reduction  will  be  effected,  if  we  divide  the  L.  C.  M. 
by  each  of  the  denominators,  and  then  multiply  the  corresponding 
numerator  by  the  quotient,  dropping  the  denominator. 

2.  The  transformation  may  be  effected  by  multiplying  each  numer- 
ator into  the  product  of  all  the  denominators  except  its  own,  omitting 
denominators. 

3.  The  transformation  may  also  be  effected  by  multiplying  both  mem- 
hers of  the  equation  hy  any  multiple  of  the  denominators. 

Exercises. 
Clear  the  following  equations  of  fractions  :  — 

1.  1  +  ^-4  =  3.  Ans.  7:r+^5a;-140  =  105. 

2.  ^  +  ^--^=:8.  Ans.  9a;  +  6a; -2a;  =  432. 

\)       u       A  i 

3.  ^  +  ^_5  +  A  =  20.     ^ns.  18a;  +  12a; -4a;  +  3a;  =720. 
2     3     9      12  ^ 

4.  ^  +  £-^  =  4.  Ans.  14a;+10a;-35a;  =  280. 

5     7     2 

5.  ^-1  +  ^=15.  Ans.  15 a; -  12a;  +  10a;  =  900. 
4     5      6 

6.  _£:=i-^I=^  =  5.  ^72s.  -2a;  +  8-a;+2  =  10. 

3  6         3 

^      '-4  =  -.  Ans.  5a;  +  60-20a;=9-3a:. 


X  5 


EQUATIONS   OF   THE    FIRST    DEGREE.  105 

8-   7-7  +  ^+^=12.     Alls.  18a:-12a:  +  9a:  +  8a;  =  864. 
4     6     8     9 

9.    -  —  -'\-f=g,  Ans.  ad  —  he -\- hdj  —  hdg , 

h         ah  d^  h'^         a 

The  L,  C.  M.  of  the  denominators  is  a^h"^. 

a*bx  -  2  a^bc'^x  +  4  a*^^  =  4  6-^2^^  _  5  ««  +  2  a^JZc^  -  3  a^b\ 

105.  Transposition  is  the  operation  of  changing  a  term  from 
one  member  to  the  other  without  destroying  the  equality  of 
the  members. 

Take,  for  example,  the  equation  5a;— 6  =  8  +  2 a;. 
If,  in  the  first  place,  we  subtract  2x  from  both  members, 
the  equality  will  not  be  destroyed,  and  we  have 

5:r  — 6  — 2a;  =  8. 

"Whence  we  see  that  the  term  2x,  which  was  additive  in  the 
second  member,  becomes  subtractive  by  passing  into  the  first. 
In  the  second  place,  if  we  add  6  to  both  members  of  the 
last  equation,  the  equality  will  still  exist,  and  we  have 

5a;- 6  — 207+6  =  8  +  6; 

or,  since  —  6  and  +  6  cancel  each  other,  we  have 

5a;  — 2.r  =  8  +  6. 

Hence  the  term  which  was  subtractive  in  the  first  member, 
'  passes  into  the  second  member  with  the  sign  of  addition. 

106.  Therefore,  for  the  transposition  of  the  terms,  we  have 
the  following  rule  :  — 

Any  term  Tuiay  he  transposed  from  one  memher  of  an  equa- 
tion to  the  other  J  if  the  sign  he  changed. 


106  ELEMENTARY    ALGEBRA. 


Exercises. 


•Transpose  the  unknown  terms  to  the  first  member,  and  the 
known  terms  to  the  second,  in  the  following :  — 

1.  Sx  +  6  —  5  =  2x—7,  Ans.  3a;  — 2a:  =  - 7  — 6  +  5. 

2.  ax-\-b  =  d  —  ex.  Ans.  ax  -\-  cx=^  d  —  b. 

3.  4a;- 3  =  2a; +  5.  Ans.  4a;-2a;  =  5  +  3. 

4.  ^x-\-  c  —  ex  —  d.  Ans.  9x  —  ex  =  —  d  —  c, 

5.  ax  +f=  dx-\-b.  Ans.  ax  —  dx  =  b—f. 

6.  6x  —  e  =  —  ax  -\-b.  Ans.  6x  +  ax  =  b  -]-  e. 

Solution  of  Equations, 

107.  The  solution  of  an  equation  is  the  operation  of  finding 
Buch  a  value  for  the  unknown  quantity  as  will  satisfy  the 
equation ;  that  is,  such  a  value  as,  being  substituted  for  the 
unknown  quantity,  will  render  the  two  members  equal.  This 
is  called  a  root  of  the  equation. 

A  root  of  an  equation  is  said  to  be  verified,  when,  being  sub- 
stituted for  the  unknown  quantity  in  the  given  equation,  the 
two  members  are  found  equal  to  each  other. 

Take  the  equation 

3a;      ^_4(a;-2)      3 
2  8 

Clearing  of  fractions  (§  104),  and  performing  the  operations 
indicated,  we  have 

12a; -32  =  4a; -8  + 24. 

Transposing  all  the  unknown  terms  to  the  first  member, 
and  the  known  terms  to  the  second  (§  106),  we  have 

12a;- 4a:  = -8  +  24  +  32. 


EQUATIONS    OF   THE    FIRST    DEGREE.  107 

Reducing  the  terms  in  the  two  members, 

8:r-48. 
Dividing  both  members  by  the  coefficient  of  x, 

48  n 

x-=  —  =  o. 
8 

Verification.     ^^ -  4  -  ii^JZ^)  _^  3^ 
2  8 

or  +9-4  =  2  +  3  =  5. 

Hence  6  satisfies  the  equation,  and  therefore  is  a  root. 

108.  By  processes  similar  to  the  above,  all  equations  of  the 
first  degree,  containing  but  one  unknown  quantity,  may  be 
solved.     Hence  the  rule  :  — 

Clear  the  equation  of  fractions^  and  perform  all  the  indi- 
cated operations. 

Transpose  all  the  unJcnown  terms  to  the  first  member,  and  all 
the  known  terms  to  the  second  member. 

Reduce  all  the  terms  in  the  first  member  to  a  single  term^ 
one  factor  of  which  shall  be  the  unknown  quantity,  and  the 
other  factor  will  be  the  algebraic  sum  of  its  coefficients. 

Divide  both  members  by  the  coefficient  of  the  unknown  quan- 
tity :  the  second  member  will  then  be  the  value  of  the  unknown 
quantity. 

Exercises. 
1.    Solve  the  equation 

5£__4£_ -j^g^T  _  13£ 
12       3  8        6  * 

Clearing  of  fractions, 

10  a;  -  32  a; -312  =  21- 52  ar. 
Transposing,  10  a;  -  32  a;  +  52  a;  =  21  +  312. 
Reducing,  30  a;  =  333. 


108  ELEMENTARY    ALGEBRA. 

Hence  ar  =  ??§  =  Ill  =  11.1, 

30       10 

a  result  which  may  be  verified  by  substituting  it  for  x  in  the  given 
equation. 

2.    Solve  the  equation 

(3a  -  x){a  -h)  +  2ax  =  ^h{x  +  a). 

Performing  the  indicated  operations,  we  have 

3  a'^  —  ao;  —  3  a5  +  hx  +  2  aa;  =  4  6a;  +  4  ah. 
Transposing, 

—  aa;  +  6a;  +  2  aa;  —  4  6a!  =  4  a&  +  3  a6  —  3  a*. 
Reducing,  aa;  —  3  6a;  =  7  a6  —  3  a^. 

Factoring,  (a  —  3  6)a;  =  7a6  —  3a^ 

Dividing  both  members  by  the  coefficient  of  a;, 

_7a6-3a^ 
a-36 

3.  Given  3a;  —  2  +  24  =  31,  to  find  x.  Ans.  rr  =  3. 

4.  Given  a;  +  18  =  3a;  —  5,  to  find  x.  Ans.  x  =  l\^, 

5.  Given   6 -2:r  + 10  =  20  -  3a;- 2,  to  find  a;. 

Ans.  x  =  2. 

6.  Given   a;  + -a;  + -a;  =  11,  to  find  a;.  Ans.  x  =  ^. 

A         o 

7.  Given   2a;-— -a;+ 1  =  5a;  — 2,  to  find  a;.         Ans.  x  =  -z' 


6-3a 


Solve  the  following  equations  :  — 

8.  3 ao;  +  -  —  3  =  5a;  —  a.                          Ans.  x      ^  ^. 

2  6a  —  26 

9.  ^^  +  ?==20-^^J^.  ^rw.  a;  =  23i. 

2          3                    2  * 

10.   ^±^  +  ^  =  4-^=^.  Ans.  x  =  Sj%. 

2        3               4  ^^ 


EQUATIONS   OF   THE   FIEST   DEGREE.  109 

11.  5_3£  +  ^^^_3.  ^^.   ^^4. 

12.  ^-2^-4=/  ^^n..  a:  =  f^/+tf' 

c  c?  -^  Zad-2bc 

13.  £zi^_2£ziM_^:^^10a+ll^. 

3  5  2 

Arts,  a; --25a +  245. 

14.  ^_§JI1^-^1^  +  11^0.  ^715.  2;  =  12. 
12        8             4          4 

15.  — ' = Ans.  X  — 


16.   8«^-5_35-£^^_^  ^,^^   ^^56  +  9^-7. 

7  2.  16a 

In  the  following  equations,  what  is  the  numerical  value  of 
X,  when  a  =  1,  5  =  2,  (?  =  3,  c?  =  4,  and  /=  6  ? 

-o     a;      X  ,  X     X      r     A  abcdf  i^« 

18. --j -=/.   Ans.  x  =  -—- — '^-— —  =  10f. 

abed  bcd—acd-\-aod  —  aoc 

19.  ^-^-^^  =  -12|f  ^ns.  a;  =  14. 

20.  :K-^i^+i^=-?  =  a;+l.  ^ns.  a;  =  6. 

io  11 

21.  a;  +  ^  +  f-^=2a;-43.  Ans,  x=m. 

4     5     6 

oo    o^     4a;  — 2     3a:  — 1  .  o 

22.  'Ax =  —7; Ans.  x  =  ?t. 

5  2 

23.  darH r —  —  x-\-a,  Ans.  x  = ^-  =  -' 

3  -^      -  6  +  5       8 


110  ELEMENTARY   ALGEBRA. 

^     ax-b  a^hx_bx-a           ^^^    ^  =  _M_=_6. 

4  3      2           3                                3a-26 

«^       ^x  20-4a;     15                                     a               oi 

25. =  —  Ans,  a:  =  3A. 

D  —  X  X                X 

29  12                        2 


2^_   («±iMzi£}_3a=Mlli!_2:.  +  ^ 


^:r 


a  —  6  a  +  6  6 

.  ^a(a^  +  3aV;  +  2ah''  -  10^-^)  _  65 

'^''  ""  b(2d'~2ab-W)  36* 

Problems  involving  Equations  of  the  First  Degree, 

109.   A  problem  is  a  question  proposed,  requiring  a  solution. 

The  solution  of  a  problem  is  the  operation  of  finding  a  quan- 
tity, or  quantities,  that  will  satisfy  the  given  conditions. 

The  solution  of  a  problem  consists  of  two  parts,  —  the  state- 
ment and  the  solution. 

The  statement  consists  in  expressing  algebraically  the  rela- 
tion between  the  known  and  the  required  quantities. 

The  solution  consists  in  finding  the  values  of  the  unknown 
quantities  in  terms  of  those  which  are  known. 

The  statement  is  made  by  representing  the  unknown  quan- 
tities of  the  problem  by  some  of  the  final  letters  of  the  alpha- 
bet, and  then  operating  upon  these  so  as  to  comply  with  the 
conditions  of  the  problem.  The  method  of  stating  problems 
is  best  learned  by  practical  examples. 

(1)  What  number  is  that  to  which  if  5  be  added  the  sum 
will  be  equal  to  9  ? 

Let  X  =  the  number. 

Then  a;  +  5  =  9. 

This  is  the  statement  of  the  problem. 

To  find  the  value  of  a:,  transpose  5  to  the  second  member. 


EQUATIONS  OF  THE  FIRST  DEGREE.         Ill 

Then  a:  =  9  -  5  =  4. 

This  is  the  solution  of  the  equation. 
Veeification.  4  +  5  =  9. 

(2)  Find  a  number  such  that  the  sum  of  one  half,  one  third, 
and  one  fourth  of  it,  augmented  by  45,  shall  be  equal  to  448. 

Let  a;  =  the  required  number. 

Then  ^  =  one  half  of  it, 

-  =  one  third  of  it, 
3 


and,  by  the  conditions, 


-  =  one  fourth  of  it, 
4 


E  +  !?  +  ^  +  45  =  448. 
2      3      4 

This  is  the  statement  of  the  problem. 

Clearing  of  fractions, 

6rc  +  4a;  +  3a;  +  540  =  5376. 

Transposing,  and  collecting  the  unknown  terms, 
13  a;  =  4836. 

Hence  x  =  ^  =  372. 

Verification. 

372  _^  372  _j_  372  _^  ^-  _  ^gg  ^  ^24  +  93  +  45  =  448. 
2         3         4 

(3)  What  number  is  that  whose  third  part  exceeds  its  fourth 

by  16? 

Let  a;  =*  the  required  number. 

Then  -  a;  =  the  third  part, 

3  ^ 

-  re  =  the  fourth  part, 


and,  by  the  conditions  of  the  problem 

1 
4 


-a;--a;=  16. 


112  ELEMENTARY   ALGEBRA. 

This  is  the  statement. 

Clearing  of  fractions,       4:X  —  Sx  =  192. 

Hence  x  =  192. 

Verification.  1^-^^e^ -is  =  16. 

3         4 

(4)  Divide  $1000  between  A,  B,  and  C  so  that  A  shall  have  * 
$72  more  than  B,  and  C  $100  more  than  A. 

Let  X  =  the  number  of  dollars  which  B  received. 

Then  x  =  B's  number, 

X  +    72  =  A's  number, 
y  (K  +  172  =  C's  number ; 

and  their  sum,  3  a;  +  244  =  1000,  the  number  of  dollars. 
This  is  the  statement. 
Transposing,  Sx==  1000  -  244  =  756. 

.'  jc  =  —  =  252  =  B's  share. 
3 

Hence  a;  +    72  =  252+    72  =  324  =  A's  share. 

a;  +  172  =  252  +  172  =  424  =  C's  share. 

Veeification.  252  +  324  +  424  =  1000. 

(5)  Out  of  a  cask  of  wine  which  had  leaked  away  a  third 
part,  21  gallons  were  afterwards  drawn,  and  the  cask,  being 
then  gauged,  was  found  to  be  half  full.  How  much  did  it 
hold? 

Let  aj  =  the  number  of  gallons. 

Then  -  =■  the  number  that  had  leaked  away, 

and  I  +  21  =  what  had  leaked  and  been  drawn. 

Hence  -  +  21  -  -. 

3  2 

This  is  the  statement. 


EQUATIONS   OF   THE    FIRST    DEGREE.  113 

Clearing  of  fractions, 

2a; +  126  =  3a;, 

and  —x  =  —  126  ; 

and  by  changing  the  signs  of  both  members,  which  does  not  destroy 
their  equality  (since  it  is  equivalent  to  multiplying  both  members  by 
—  1),  we  have 

X  =  126. 

Verification.      —  +  21  =  42  h-  21  =  63  =  —• 
3  2 

(6)  A  fish  was  caught  whose  tail  weighed  9  pounds.  His 
head  weighed  as  much  as  his  tail  and  half  his  body,  and  his 
body  weighed  as  much  as  his  head  and  tail  together.  What 
was  the  weight  of  the  fish  ? 

Let  2  a;  =  the  weight  of  the  body  in  pounds. 

Then  9  +  a;  =  weight  of  the  head  ; 

and,  since  the  body  weighed  as  much  as  both  head  and  tail, 

2a;  =  9  +  9  +  a;, 
which  is  the  statement. 

Then  2a;  -  a;  =  18,  and  a;  =  18. 

Hence  we  have    2  a;  =  36  lbs.  =  weight  of  the  body, 
9  +  a;  =  27  lbs.  =  weight  of  the  head, 
9  lbs.  =  weight  of  the  tail. 
Hence  72  lbs.  =  weight  of  the  fish. 

(7)  The  sum  of  two  numbers  is  67,  and  their  difference  19. 
What  are  the  two  numbers  ? 

First  Method. 

Let  X  =  the  less  number. 

Then  a;  +  19  =  the  greater, 

and  2a; +  19  =  67. 

This  is  the  statement. 

D.  N.  E.  A.  —  S. 


114  ELEMENTARY    ALGEBRA. 


Transposing, 

2a;  =  67 -19  =  48. 

Hence 

a;  =  ^  =  24,  and  a;  +  19  =  43. 

Verification. 

43  +  24  =  67,  and  43  -  24  =  19. 

Second  Method. 

Let 

Then               x - 
id                     2a;- 

X  =  the  greater  number. 

-  19  =  the  less, 

-  19  =  67 ;  whence  2a;  =  67  +  19. 

2 

nsequently       x  - 

-19  =  43-19  =  24. 

As  a  general  solution  of  this  problem,  take  the  following :  — 
The  sum  of  two  numbers  is  s.    Their  difference  is  d.   What 
are  the  two  numbers  ? 

Let  X  =  the  less  number. 

Then  x  -i-  d  =  the  greater, 

and  2x  +  d  =  s,  their  sum. 

TTTv  s  —  d     s      d 

Whence  x  = = ; 

2         2     2 

and  consequently  x  +  d  = [■  d  =  -  -\ — 

^         ^  2      2  2      2 

As  these  two  results  are  not  dependent  on  particular  values 
attributed  to  s  or  d,  it  follows  that 

The  greater  of  two  numbers  is  equal  to  half  their  sum,  plus 
half  their  difference. 

The  less  is  equal  to  half  their  sum,  minus  half  their  differ- 
ence. 

Thus,  if  the  sum  of  two  numbers  is  32,  and  their  difference 
16,  the  greater  is 

the  less,  §2_16_jg_g^g 

2       2 
Verification.    24  +  8  =  32,  and  24  -  8  =  16. 


EQUATIONS  OF  THE  FIRST  DEGREE.         115 

(8)  A  person  engaged  a  workman  for  48  days.  For  each 
day  that  he  labored  he  received  24  cents,  and  for  each  day 
that  he  was  idle  he  paid  12  cents  for  his  board.  At  the  end 
of  the  48  days  the  account  was  settled,  when  the  laborer 
received  504  cents.  Required  the  number  of  working  days, 
and  the  number  of  days  he  was  idle. 

If  the  number  of  working  days  and  the  number  of  idle  days  were 
known,  and  the  first  multiplied  by  24  and  the  second  by  12,  the  difi'er- 
ence  of  these  products  would  be  504.  Let  us  indicate  these  operations 
by  means  of  algebraic  signs. 

Let  X  =  the  number  of  working  days. 

Then  48  —  a;  =  the  number  of  idle  days, 

24  X  a?  =  the  amount  earned, 
and  12  (48  —  x)  =  the  amount  paid  for  board. 

Then       24  a;  -  12  (48  -  a:)  5=  504, 
what  was  received,  which  is  the  statement. 

Then,  performing  the  operations  indicated, 
24  a; -576  + 12  a;  =  504, 
or  36  a;  =  504 +  576  =  1080, 

and  a;  =  -— -  =  30,  the  number  of  working  days ; 

00 

whence  48  -  30  =  18,  the  number  of  idle  days. 

Verification. 

30  days'  labor  at  24  cents  =  30  x  24  =  720  cents. 
18  days'  board  at  12  cents  =  18x12  =  216  cents. 

Difference,  or  amount  received  =  504  cents. 

This  problem  may  be  made  general  by  denoting  the  whole 
number  of  working  and  idle  days  by  n;  the  amount  received 
for  each  day's  work,  by  a;  the  amount  paid  for  board  for 
each  idle  day,  by  b ;  and  what  was  due  the  laborer,  or  the 
balance  of  the  account,  by  c. 

As  before,  let  the  number  of  working  days  be  denoted  by  x. 
The  number  of  idle  days  will  then    be  denoted   by   n  —  x. 


116  ELEMENTARY   ALGEBRA. 

Hence  what  is  earned  will  be  expressed  by  ax ;  and  the  sum 
to  be  deducted  on  account  of  board,  by  b{n  —  x). 
The  statement  of  the  problem,  therefore,  is, 

ax  —  h  {n  —  x)  =^  c. 

Performing  the  indicated  operations, 

ax  —  hn  +  hx  =  c,  or  (a  +  6)  re  =  c  +  hn. 

Whence  x  = —  =  number  of  working  days  ; 

and  n-a;==:n-^  +  ^^  =  ^^-^^^-^-^^ 

=  number  of  idle  days. 


a  +  6 


Let  US  suppose  n  ^  48,  a  :=  24,  h  =  12,  and  c  =  504. 
These  numbers  will  give  for  x  the  same  value  as  before  found. 

(9)  A  person,  dying,  leaves  half  of  his  property  to  his  wife, 
one  sixth  to  each  of  two  daughters,  one  twelfth  to  a  servant, 
and  the  remaining  $600  to  the  poor.  What  was  the  amount 
of  the  property  ? 

Let  X  =  the  amount  in  dollars. 

Then  -  =  what  he  left  to  his  wife, 

2 

-  =  what  he  left  to  one  daughter, 

and  -—  =  -,  what  he  left  to  both  daughters-, 

6       3 

also  —  =  what  he  left  to  his  servant, 

12 

and  $600  =  what  he  left  to  the  poor. 

Then,  by  the  conditions, 

E  ^  i?  -j.  JL  -I-  600  =  X,  the  amount  of  the  property, 
2     3      12  t    f     J> 

which  gives  x  =  $7200. 


EQUATIONS   OF    THE    FIRST    DEGREE.  117 

(10)  A  and  B  play  together  at  cards.  A  sits  down  with 
$84,  and  B  with  $48.  Each  loses  and  wins  in  turn,  when  it 
appears  that  A  has  five  times  as  much  as  B.  How  much  did 
A  win  ? 

Let  X  denote  the  number  of  dollars  A  won. 

Then  A  rose  with  84  +  a;  dollars, 

and  B  rose  with  48  —  a;  dollars. 

But  by  the  conditions  we  have 

84  +  a;  =  5(48-ir). 

Hence  84  +  a;  =  240  -  5  a;, 

and  6  a;  =  156. 

Consequently  x  =  26,  or  A  won  |26. 

Verificatioit. 

84  +  26  =  110,  48  -  26  =  22,  110  =  5  (22)  =  110. 

(11)  A  can  do  a  piece  of  work  alone  in  10  days,  and  B  in 
13  days.     In  what  time  can  they  do  it  if  they  work  together  ? 

Denote  the  time  by  x,  and  the  work  to  be  done  by  1.     Then 

In"  1  day   A  can  do  —  of  the  work, 

•^  10  '  , 

and  B  can  do  —  of  the  work  : 

13 

and  In  x  days  A  can  do  -^  of  the  work, 

^  10 

and  B  can  do  -^  of  the  work. 

Hence,  by  the  conditions, 

^  +  ii  =  1,  which  gives  13  a;  +  lOx  =  130. 
'       Hence  23  a;  =  130,   a;  =  —  =  5Jf  days. 

(12)  A  fox,  pursued  hy  a  hound,  has  a  start  of  60  of  his 
own  leaps.  Three  leaps  of  the  hound  are  equivalent  to  7  of 
the  fox ;  but  while  the  hound  makes  6  leaps,  the  fox  makes  9. 
How  many  leaps  must  the  hound  make  to  overtake  the  fox  ? 


118  ELEMENTARY    ALGEBRA. 

There  is  some  difficulty  in  this  problem,  arising  from  the  different 
units  which  enter  into  it. 

Since  3  leaps  of  the  hound  are  equal  to  7  leaps  of  the  fox,  1  leap  of 

'7 

the  hound  is  equal  to  7-  fox  leaps. 
o 

Since,  while  the  hound  makes  6  leaps,  the  fox  makes  9,  while  the 

9        3 

hound  makes  1  leap,  the  fox  will  make  -,  or  -,  leaps. 

6        2 
Let  X  denote  the  number  of  leaps  which  the  hound  makes  before  he 
overtakes  the  fox,  and  let  1  fox  leap/ienote  the  unit  of  distance. 

Since  1  leap  of  the  hound  is  equal  to  -  of  a  fox  leap,  x  leaps  will  be 

7  "^ 

equal  to  -a;  fox  leaps  ;  and  this  will  denote  the  distance  passed  over  by 

the  hound  in  fox  leaps. 

Since,  while  the  hound  makes  1  leap,  the  fox  makes  -  leaps,  while  the 

3  -^ 

hound  makes  x  leaps,  the  fox  makes  -x  leaps ;  and  this  added  to  60,  his 

o  2 

distance  ahead,  will  give  -x  +  60  for  the  whole  distance  passed  over  by 


le  fox. 

^ 

Hence,  from  the  conditions. 

2.= 

3 

-  fx  +  60. 

Whence 

Ux  = 

=  9a; +  360, 

X  = 

=  72. 

The  hound,  therefore,  makes  72  leaps  before  overtaking  the  fox.     In 
the  same  time,  the  fox  makes  72  x  7  =  108  leaps. 


Verification.      108  -f  60  =  168,  whole  number  of  fox  leaps, 

.7 
3" 


!X-  =  1G8. 


Exercises. 

1.  A  father  leaves  his  property,  amounting  to  $2520,  to 
four  sons.  A,  B,  C,  and  D.  C  is  to  have  $360 ;  B,  as  much  as 
C  and  D  together ;  and  A,  twice  as  much  as  B,  less  $1000. 
How  much  do  A,  B,  and  D  receive  ? 

Ans.  A,  $760 ;  B,  $880 ;  D,  $520. 


EQUATIONS    OF    THE    FIRST    DEGREE.  119 

2.  An  estate  of  $7500  is  to  be  divided  among  a  widow, 
two  sons,  and  three  daughters,  so  that  each  son  shall  receive 
twice  as  much  as  each  daughter,  and  the  widow  herself  $500 
more  than  all  the  children.  What  was  her  share,  and  what 
the  share  of  each  child  ?  r  Widow's  share,     $4000. 

Ans.  }  Each  son's,  $1000. 

(  Each  daughter's,    $500. 

3.  A  company  of  180  persons  consists  of  men,  women,  and 
children.  The  men  are  8  more  in  number  than  the  women, 
and  the  children  20  more  than  the  men  and  women  together. 
How  many  of  each  sort  in  the  company  ? 

Ans.  44  men,  36  women,  100  children. 

4.  A  father  divides  $2000  among  five  sons,  so  that  each 
elder  should  receive  $40  more  than  his  next  younger  brother. 
What  is  the  share  of  the  youngest  ?  Ans.  $320. 

5.  A  purse  of  $2850  is  to  be  divided  among  three  persons, 
A,  B,  and  0.  A's  share  is  to  be  to  B^s  as  6  to  11,  and  C  is  to 
have  $300  more  than  A  and  B  together.  What  is  each  one's 
share?  Ans.  A's,  $450;  B"s,  $825 ;  C's,  $1575. 

6.  Two  pedestrians  start  from  the  same  point  and  travel  in 
the  same  direction.  The  first  steps  twice  as  far  as  the  second ; 
but  the  second  makes  5  steps  while  the  first  makes  but  1. 
At  the  end  of  a  certain  time  they  are  300  feet  apart.  Now, 
allowing  each  of  the  longer  paces  to  be  3  feet,  how  far  will 
each  have  traveled?  Ans.  1st,  200  feet ;  2d,  500  feet. 

7.  Two  carpenters,  24  journeymen,  and  8  apprentices 
received  at  the  end  of  a  certain  time  $144.  The  carpenters 
received  $1  per  day;  each  journeyman,  half  a  dollar;  and 
each  apprentice,  25  cents.  How  many  days  were  they  em- 
ployed? An$.  9  days. 


120  ELEMENTARY    ALGEBRA. 

8.  A  capitalist  receives  a  yearly  income  of  $2940.  Four 
fifths  of  his  money  bears  an  interest  of  4  per  cent,  and  the 
remainder  5  per  cent.     How  much  has  he  at  interest? 

Ans.  $70,000. 

9.  A  cistern  containing  60  gallons  of  water  has  three  une- 
qual cocks  for  discharging  it.  The  largest  will  empty  it  in 
one  hour ;  the  second,  in  two  hours ;  and  the  third,  in  three. 
In  what  time  will  the  cistern  be  emptied  if  they  all  run 
together  ?  Ans.  32y^y  minutes. 

10.  In  a  certain  orchard  one  half  are  apple  trees ;  one 
fourth,  peach  trees ;  one  sixth,  plum  trees.  There  are  also 
120  cherry  trees  and  80  pear  trees.  How  many  trees  in  the 
orchard?  Ans.  2400. 

11.  A  farmer,  being  asked  how  many  sheep  he  had,  an- 
swered that  he  had  them  in  five  fields.  In  the  first  he  had 
\]  in  the  second,  ^',  in  the  third,  -|- ;  in  the  fourth,  y^-^-;  and 
in  the  fifth,  450.     How  many  had  he  ?  Ans.  1200. 

12.  My  horse  and  saddle  together  are  worth  $132,  and  the 
horse  is  worth  ten  times  as  much  as  the  saddle.  What  is  the 
value  of  the  horse  ?  Ans.  $120. 

13.  The  rent  of  an  estate  is  this  year  8  per  cent  greater 
than  it  was  last.  This  year  it  is  $1890.  What  was  it  last 
year?  Ans.  $1750. 

14.  What  number  is  that  from  which,  if  5  be  subtracted, 
■|  of  the  remainder  will  be  40  ?  Ans.  65. 

15.  A  post  is  }  in  the  mud,  i  in  the  water,  and  10  feet 
above  the  water.     What  is  the  whole  length  of  the  post? 

Ans.  24  feet. 

16.  After  paying  \  and  ^  of  my  money,  I  had  66  guineas 
left  in  my  purse.     How  many  guineas  were  in  it  at  first? 

A71S.  120. 


EQUATIONS    OF   THE    FIRST    DEGREE.  121 

17.  A  person  was  desirous  of  giving  3  pence  apiece  to  some 
beggars,  but  found  he  had  not  money  enough  in  his  pocket 
by  8  pence.  He  therefore  gave  them  each  2  pence,  and  had  3 
pence  remaining.    Required  the  number  of  beggars.    Ans.  11. 

18.  A  person,  in  play,  lost  J  of  his  money,  and  then  won 
3  shillings,  after  which  he  lost  J  of  what  he  then  had,  and, 
this  done,  found  that  he  had  but  12  shillings  remaining. 
What  had  he  at  first  ?  Ans.  20  shillings. 

19.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money 
in  trade.  A  gains  $126,  and  B  loses  $87,  and  A's  money  is 
then  double  B's.     What  did  each  lay  out?  A7is.  $300. 

20.  A  person  goes  to  a  tavern  with  a  certain  sum  of  money 
in  his  pocket,  where  he  spends  2  shillings.  He  then  borrows 
as  much  money  as  he  had  left,  and,  going  to  another  tavern, 
he  there  spends  2  shillings  also.  Then  borrowing  again  as 
much  money  as  was  left,  he  went  to  a  third  tavern,  where 
likewise  he  spent  2  shillings,  and  borrowed  as  much  as  he  had 
left;  and,  again  spending  2  shillings  at  a  fourth  tavern,  he 
then  had  nothing  remaining.     What  had  he  at  first  ? 

Ans,   3s.  9d. 

21.  A  tailor  cut  19  yards  from  each  of  three  equal  pieces 
of  cloth,  and  17  yards  from  another  of  the  same  length,  and 
found  that  the  four  remnants  were  together  equal  to  142 
yards.     How  many  yards  in  each  piece  ?  Ans.  64. 

22.  A  fortress  is  garrisoned  by  2600  men,  consisting  of 
infantry,  artillery,  and  cavalry.  Now,  there  are  nine  times 
as  many  infantry,  and  three  times  as  many  artillery,  as  there 
are  cavalry.     How  many  are  there  of  each  corps  ? 

Ans.  200  cavalry,  600  artillery,  1800  infantry. 

HP  23.  All  the  journeyings  of  an  individual  amounted  to  2970 
miles.     Of  these,  he  traveled  3i  times  as  many  by  water  as 


l'2'2  ELEMENTARY    ALGEBRA. 

on  horseback,  and  2i  times  as  many  on  foot  as  by  water. 
How  many  miles  did  he  travel  in  each  way  ? 

Ans.  240  miles,  840  miles,  1890  miles. 

24.  A  sum  of  money  was  divided  between  two  persons,  A 
and  B.  A's  share  was  to  B's  in  the  proportion  of  5  to  3,  and 
exceeded  five  ninths  of  the  entire  sum  by  $50.  What  was 
the  share  of  each?  Ans.  A's  share,  |450 ;  B's,  $270. 

25.  Divide  a  number  a  into  three  such  parts  that  the 
second  shall  be  n  times  the  first,  and  the  third  m  times  as 
great  as  the  first. 

Ans.  1st, ;    2d, "^ ;    3d, 


1-i-m  +  n'        '1  +  771  +  n'        '  l-\-m-\-n 

26.  A  father  directs  that  $1170  shall  be  divided  among  his 
three  sons  in  proportion  to  their  ages.  The  oldest  is  twice  as 
old  as  the  youngest,  and  the  second  is  one  third  older  than 
the  youngest.     How  much  was  each  to  receive  ? 

Ans,  Youngest,  $270 ;  second,  $360;  oldest,  $540. 

27.  Three  regiments  are  to  furnish  594  men,  and  each  to 
furnish  in  proportion  to  its  strength.  Now,  the  strength  of 
the  first  is  to  the  strength  of  the  second  as  3  to  5 ;  and  that 
of  the  second  to  that  of  the  third  as  8  to  7.  How  many  must 
each  furnish  ? 

Ans.  1st,  144  men  ;  2d,  240  men  ;  3d,  210  men. 

28.  Five  heirs.  A,  B,  C,  D,  and  E,  are  to  divide  an  inheri- 
tance of  $5600.  B  is  to  receive  twice  as  much  as  A,  and  $200 
more ;  C,  three  times  as  much  as  A,  less  $400 ;  D,  the  half  of 
what  B  and  C  receive  together,  and  $150  more ;  and  E,  the 
fourth  part  of  what  the  four  others  get,  plus  $475.  How  • 
much  did  each  receive  ? 

Ans.  A,  $500;  B,$1200;  C,$1100;  D,$1300;  E,$1500. 

29.  A  person  has  four  casks,  the  second  of  which  being 
filled  from  the  first,  leaves  the  first  four  sevenths  full;  the 


EQUATIONS  OF  THE  FIRST  DEGREE.         123 

third  being  filled  from  the  second,  leaves  it  one  fourth  full ; 
and  when  the  third  is  emptied  into  the  fourth,  it  is  found  to 
fill  only  nine  sixteenths  of  it.  But  the  first  will  fill  the  third 
and  fourth,  and  leave  15  quarts  remaining.  How  many 
gallons  does  each  hold  ? 
Ans.  1st,  35  gals. ;  2d,  15  gals. ;  3d,  Hi  gals. ;  4th,  20  gals. 

30.  A  courier,  having  started  from  a  place,  is  pursued  by 
a  second  after  the  lapse  of  10  days.  The  first  travels  4  miles 
a  day  ;  the  other,  9.  How  many  days  before  the  second  will 
overtake  the  first?  Ans.  8  days. 

31.  A  courier  goes  31i  miles  every  five  hours,  and  is  fol- 
lowed by  another  after  he  had  been  gone  eight  hours.  The 
second  travels  22J  miles  every  three  hours.  How  many  hours 
before  he  will  overtake  the  first  ?  Ans.  42  hours. 

32.  Two  places  are  80  miles  apart,  and  a  person  leaves  one 
of  them  and  travels  towards  the  other  at  the  rate  of  3  J  miles 
per  hour.  Eight  hours  after,  a  person  departs  from  the  second 
place,  and  travels  at  the  rate  of  5^  miles  per  hour.  How  long 
before  they  will  be  together  ?  "  Ans.  6  hours. 

Equations  containing  Two  Unhnoion  Quantities. 

110.    If  we  have  a  single  equation,  as 

2.r  +  32/-21, 

containing  two  unknown  quantities,  x  and  y,  we  may  find  the 
value  of  one  of  them  in  terms  of  the  other ;  as 


21-3 


1  (1) 


2 

Now,  if  the  value  of  y  is  unknown,  that  of  x  will  also  be 
unknown.  Hence,  from  a  single  equation  containing  two  un- 
known quantities,  the  value  of  x  cannot  be  determined. 


124  ELEMENTARY    ALGEBRA. 

If  we  have  a  second  equation,  as 

we  may,  as  before,  find  the  value  of  x  in  terms  of  3/,  giving 

X  = -— ^  (2) 

t  ^ 

Now,  if  the  values  of  x  and  y  are  the  same  in  Equations  (1) 

and  (2),  the  second  members  may  be  placed  equal  to  each 

other,  giving 

21-3y^35  — 4y 

2    '  5      ' 

or  105-15y-:70-8y, 

from  which  we  find    y  =  5. 

Substituting  this  value  for  y  in  Equations  (1)  or  (2),  we 
find  x  =  3.  Such  equations  are  called  simultaneous  equations. 
Hence 

111,  Simultaneous  equations  are  those  in  which  the  values  of 
the  unknown  quantities  are  the  same  in  both. 

Elimination. 

112.  Elimination  is  the  operation  of  combining  two  equa- 
tions containing  two  unknown  quantities,  and  deducing  there- 
from a  single  equation  containing  but  one. 

There  are  three  principal  methods  of  elimination, — 
I.    By  addition  or  subtraction. 
II.    By  substitution. 
.  III.    By  comparison. 
We  shall  cpnsider  these  methods  separately. 

I.   Elimination  by  Addition  or  Subtraction. 

(1)  Take  the  two  equations 

Zx-2y=-    7, 
8a:4-2y=-48. 


EQUATIONS    OF   THE    FIRST   DEGREE.  125 

If  we  add  these  two  equations,  member  to  member,  we 

obtain 

ll:r  =  55, 

which  gives,  by  dividing  by  11, 

x  =  5; 

and  substituting  this  value  in  either  of  the  given  equations, 

we  find 

(2)  Again,  take  the  equations 

8^ +  23/ =  48, 

Sx  +  2y  =  2S. 

If  we  subtract  the  2d  equation  from  the  1st,  we  obtain 

5:r-:25, 
which  gives,  by  dividing  by  5, 

X  =  5  ; 
and  by  substituting  this  value,  we  find 

y-4. 

(3)  Given  the  sum  of  two  numbers  equal  to  s,  and  their 
difference  equal  to  c?,  to  find  the  numbers. 

Let  X  =  the  greater,  and  y  the  less,  number. 

Then  x  +  y  =  s, 

and  x  —  y  =  d. 

By  adding  (J  102,  Axiom  1),  2x  =  s  -{-  d. 
By  subtracting  (§  102,  Axiom  2), 

2y  =  s  —  d. 
Each  of  these  equations  contains  but  one  unknown  quantity. 

From  the  first  we  obtain         x  = ; 

2 

s  —  d 
and  from  the  second,  y  = 

A 

Note.  —  These  are  the  same  values  as  were  found  in  Problem  (7), 
p.  114. 


126  ELEMENTARY   ALGEBRA. 

(4)  A  person  engaged  a  workman  for  48  days.  For  each 
day  that  he  labored  he  was  to  receive  24  cents,  and  for  each 
day  that  he  was  idle  he  was  to  pay  12  cents  for  his  board. 
At  the  end  of  the  48  days  the  account  was  settled,  when  the 
,  laborer  received  504  cents.  Required  the  number  of  working 
days,  and  the  number  of  days  he  was  idle. 

Let  X  =  the  number  of  working  days, 

and  y  =  the  number  of  idle  days. 

Then  24  aj  =  what  he  earned, 

and  123/ =  what  he  paid  for  his  board. 

Then,  by  the  conditions  of  the  question,  we  have 
a;  +  2/  =  48, 
and  24  a; -12?/ =  504. 

This  is  the  statement  of  the  problem. 

It  has  already  been  shown  (§  102,  Axiom  3)  that  the  two  members 
of  an  equation  may  be  multiplied  by  the  same  number  without  destroy- 
ing the  equality.  Let,  then,  the  members  of  the  first  equation  be 
multiplied  by  24,  the  coefficient  of  a;  in  the  second.     We  shall  then  have 

24a; +  243/ =  1152 
24a;-12.v=    504 

and,  by  subtracting,         263/=    648 

648      .0 
^       36 

Substituting  this  value  of  3/  in  the  equation 

24  a; -123/ =  504, 

we  have  24  a;  -  216  =  504, 

which  gives  24  a;  =  504  +  216  =  720, 

A  720     o^ 

and  X  =»  -—-  =  30. 

24 

(See  p.  115,  where  this  problem  is  solved  by  means  of  one  unknown 
quantity.) 


EQUATIONS   OF   THE    FIRST    DEGREE. 


127 


113.  In  a  similar  manner,  either  unknown  quantity  may 
be  eliminated.     Hence  the  following  rule  :  — 

Prepare  the  equations  so  that  the  coefficients  of  the  quantity 
to  be  eliminated  shall  be  numerically  equal. 

If  the  signs  are  unlike,  add  the  equations,  member  to  m^em- 
ber ;  if  alike,  subtract  them,  member  from  member. 


Exercises. 


Find  the  values  of  x  and  y,  by  addition  or  subtraction,  in 
the  following  simultaneous  equations  :  — 


1. 


3. 


5. 


7. 


|3a;-    y-3.) 
1    y  +  2x==1.) 

(4a;-7y-:-22.  ) 
l5a;  +  2y---37.      j 

|2a;+6y  =  42. 1 

I  8a; -9y-l.    ) 

(1407 -157/ -12.") 
(    lx+   83/  =  37.l 


--a;  +  -v=  6. 

2        3 


1,1  A 

7        8^ 
X-    2/ =  -2. 


Ans.  x  =  2,  y  =  S. 

Ans.  x  =  5,  y  =  ^. 

Ans.  x  =  Ai,  y  =  5|-. 


A  1  1 

Ans.  x  =  -.    3/=-. 


^^25.  o;  —  3,  y  —  2. 


Ans.  x  =  Q,  3/  =  9. 


Ans.   a:  =  14,  y=  16. 


128  ELEMENTARY    ALGEBRA. 

8.  Says  A  to  B,  "  You  give  me  $40  of  your  money,  and  I 
shall  then  have  five  times  as  much  as  you  will  have  left." 
Now,  they  both  had  $120.    How  much  had  each?     Ans.  $60. 

9.  A  father  says  to  his  son,  "  Twenty  years  ago  my  age 
was  four  times  yours:  now  it  is  just  double."  What  were 
their  ages  ?  Ans.  Father's,  60  years  ;  son's,  30  years. 

10.  A  father  divided  his  property  between  his  two  sons. 
At  the  end  of  the  first  year  the  elder  had  spent  one  quarter 
of  his,  and  the  younger  had  made  $1000,  and  their  property 
was  then  equal.  After  this,  the  elder  spent  $500,  and  the 
younger  made  $2000,  when  it  appeared  that  the  younger  had 
just  double  the  elder.     What  had  each  from  the  father? 

Ans.  Elder,  $4000 ;  younger,  $2000. 

11.  If  John  gives  Charles  15  apples,  they  will  have  the 
same  number  ;  but  if  Charles  gives  15  to  John,  John  will  have 
15  times  as  many,  wanting  10,  as  Charles  will  have  left.  How 
many  has  each  ?  Ans.  John,  50 ;  Charles,  20. 

12.  Two  clerks,  A  and  B,  have  salaries  which  are  together 
equal  to  $900.  A  spends  -^^  per  year  of  what  he  receives, 
and  B  adds  as  much  to  his  as  A  spends.  At  the  end  of  the 
year  they  have  equal  sums.     What  was  the  salary  of  each  ? 

Ans.  A's,  $500 ;  B's,  $400. 

II.   Elimination  by  Substitution. 
114.   Let  us  take  the  equations 

5^+ 7?/ =  43  (1) 

Ux  +  9i/  =  69  (2) 

Find  the  value  of  x  in  the  first  equation,  which  gives 

.  =  43-7^. 


EQUATIONS    OF    THE    FIRST    DEGREE.  129 

Substitute  this  value  of  x  in  the  second  equation,  and  we 
have 

or  473-77y  +  45y  =  345, 

or  -32y  =  -128. 

Here  x  has  been  eliminated  by  substitution. 
In  a  similar  manner  we  can  eliminate  any  unknown  quan- 
tity.    Hence  the  rule  :  — 

Find  from  either  equation  the  value  of  the  unknown  quan- 
tity to  be  eliminated. 

Substitute  this  value  for  that  quantity  in  the  other  equation. 

Note.  —  This  method  of  elimination  is  used  to  great  advantage  when 
the  coefficient  of  either  of  the  unknown  quantities  is  1. 

Exercises. 

Find  by  the  last  method  the  values  of  x  and  y  in  the  fol- 
lowing equations :  — 

1.  ?>x  —  y  —  \,  and  3y  — 22:  =  4.  Ans.  x  =  \,  3/ =  2. 

2.  5y-4a;  =  -22,  and  3y  +  4:r-^38.  Ans.  rr-S,  y=^2, 

3.  5:  +  8y  =  18,  andy-3a;  =  -29.        Ans.  x  =  10,  y  =1. 

4.  5^-y-:13,  and8a;  +  ?y==29.      Ans.  x  =  ?>i,  y  =--Ah. 

9 

5.  10.r-|=69,  and  10y-^'  =  49.         Ans.  x^l,  y  =  b. 

6.  ^  +  la:--^=10,  and  "  +  :^  =  2.      Ans.  x  =  ^,  y  =  10. 

2        5  8      10  ^ 

7.  1-^+5  =  2,  and:r  +  |  =  17f       Ans,  x  =  lb,  y  =  IL 
7      3  5 

D.  N.E.A.  —  9. 


130  ELEMENTARY    ALGEBRA. 


8.  |  +  |+3  =  6i,and|-^==|.       Arts,  x  :=  Sh  y  =  i. 

9.  8~!+^^^'^'''^5~ft^^-   ^^^-^  =  12,^=16. 

10.  ^-^_1=_9,  and5a;-^-29. 
7       2  49 

Ans.  x  =  6,  y  =  7. 

11.  Two  misers,  A  and  B,  sit  down  to  count  over  their 
money.  Together  they  have  $20,000,  and  B  has  three  times 
as  much  as  A.     How  much  has  each  ? 

Ans.  A,  $5000 ;  B,  $15,000. 

12.  A  person  has  two  purses.  If  he  puts  $7  into  the  first, 
the  whole  is  worth  three  times  as  much  as  the  second  purse  ; 
but  if  he  puts  $7  into  the  second,  the  whole  is  worth  five 
times  as  much  as  the  first.     What  is  the  value  of  each  purse  ? 

Ans.  1st,  $2 ;  2d,  $3. 

13.  Two  numbers  have  the  following  relations  :  if  the  first 
be  multiplied  by  6,  the  product  will  be  equal  to  the  second 
multiplied  by  5  ;  and  1  subtracted  from  the  first  leaves  the 
same  remainder  as  2  subtracted  from  the  second.  What  are 
the  numbers  ?  Ans.  5  and  6. 

14.  Find  two  numbers  with  the  following  relations :  the 
first  increased  by  2  is  3}  times  as  great  as  the  second ;  and 
the  second  increased  by  4  gives  a  number  equal  to  half  the 
first.  Ans.  24  and  8. 

15.  A  father  says  to  his  son,  "  Twelve  years  ago  I  was 
twice  as  old  as  you  are  now  ;  four  times  your  age  at  that 
time,  plus  twelve  years,  will  express  my  age  twelve  years 
hence."     What  were  their  ages  ? 

Ans.  Father,  72  years  ;  son,  30  years. 


EQUATIONS  OF  THE  FIRST  DEGEEE.         13r 

III.   Elimination  by  Comparison, 
116.    Take  again  the  equations, 

5a:  +  7y  =  43, 
lla;  +  9y  =  69. 

Finding  the  value  of  x  from  the  first  equation,  we  have 
43 -7y 
"^— 5        ' 

and  finding  the  value  of  x  from  the  second,  we  obtain 

69  — 9y 

Let  these  two  values  of  x  be  placed  equal  to  each  other, 
and  we  have 

43-7y_69-9y 
5    '  11      ' 

or  473  -  77y  =  345  -  45y, 

or  -32y  =  ~128. 

Hence  y  =  4, 

1                                               69  -  36      o 
and  X  = =  3. 

This  method  of  elimination  is  called  the  method  by  com- 
parison, for  which  we  have  the  following  rule  :  — 

Mnd  from  each  equation  the  value  of  the  same  unknown 
quantity/  to  he  eliminated. 

Place  these  values  equal  to  each  other. 

Exercises. 

Find  by  the  last  rule  the  values  of  x  and  y  from  the  follow- 
ing equations :  — 

1.   3a;  +  |+6  =  42,  andy-^-H^. 

Ans.  a;=  11,  y  =  15. 


132  ELEMENTARY    ALGEBRA. 

2.  '^-f+5  =  6,  and  f +  4  =  ^+6. 
4      7  5  14 

Ans.  a:  =^  28,  y  =  20. 

?/       a:      22 

3.  ^~7  +  ~^~l)  ^"^  3y--a;  — 6.       ^ns.  :r  ^  9,  y  =  5. 

4.  y-3-^a;+5,  and^±^-y-3f   ^ns.  :r  =  2,  y  -  9. 

6.  2^-^+1  =  ^-2,  and  1  +  1  =  ^-13. 

J^ns.  a:  =  16,  y  =  7. 

6-   ^^  +  ^^-=^^-^^,  and  a;  +  y-16. 
2  2  3  "^ 

-47ZS.  X  — ■  10,  ^  =^Q. 

7.  ±£:^l^  =  a;-2|,   a:-'^!-=i-:0.       ^7i5.  a:=l,  y  =  3. 


8.    2y  +  3a;-y  +  43,  y-^=^y-| 


9.    4y-— -^  =  a:+18,  and  27-y-a;  +  y4-4 
A 


Ans.  a;  =  10,  y—  13. 
Ans.  a;  =  9,  y  =  ^^ 


10.   i-^iz:^  +  4-y-16i  ^ 

^n5.  a:  =  10,  y  =  20. 


116.  Having  explained  the  principal  methods  of  elimina- 
tion, we  shall  add  a  few  examples  which  may  be  solved  by 
any  one  of  them ;  and  often,  indeed,  it  may  be  advantageous 
to  employ  them  all,  even  in  the  same  example. 


EQUATIONS    OF    THE    FIRST    DEGREE. 


133 


General  Exercises. 

Find  the  values  of  x  and  y  in  the  following  simultaneous 
equations :  — 

1.    2x  +  Si/  =  l6,  and  Sx-2'i/=ll.      Arts.  x  =  5,  y  =  2. 

^     2x  .  Sy       9  .    3a;  ,  2y       61 

2. — ^  =  — ,  and    ■ — —  = 

5^4       20'  4^5       120 

Ans.  x  =  -,  y  —  -' 
2    "^      3 

3.    -+7y.=.99^  and  ^+7a:  =  51.  Ans.  x-=1,  y  =  \\, 

7  7 

'2  4^    '      5      '  3  4      ^      * 

An8.  x  =  60,  ?/  =  40. 

X  —  G. 


6. 


"^-2^  +  ^^^^' 


7:r  =  41. 


6.     ^ 


2 

T 


1 


^-y  +  4iy-12i. 


7.     < 


'  Zy-x     2x~y _. 
6      "^      4 

8—  2r 


-4  ns. 


^?2S.   - 


Ans. 


8. 


^3^— 8,y  —  6,  TQ7 

-^-+^  +  y-^i8TV 


8a;-3- 


6- 


^^79. 


y  =  8. 

a;  =  5. 

y  =  3. 
a;  =  9. 

y  =  8. 

a;  =  10. 

y  =  l2. 


134 


ELEMENTARY    ALGEBRA. 


'  4a; -4     .v-5  ,  ^__ 


+  6-12|. 


9.     < 


10. 


1  1      ,  2/-4     5 

2  3^^     3         3 


ax  —  by  =  c. 


Ans. 


X  —  6. 


a  —  y-\-x  =  d. 


Ans. 


l^x  +1y-  341  -  7Jy  +  43irr. 


\ 


y  =  b, 

c  -{-  ah  —  hd 
a  —  h 

a^  +  c  —  ad 
a  —  h 

a;  =  -  12. 


Ans. 


'^-     l2.;+10^3y+L  I     ^^-'•|y^-5. 


13.     < 


14. 


15.      -^ 


16. 


ax  =  hy. 

x  +  y==^c. 
ax  +  hy  ^=  c. 

fx  +  gy  =  h. 

a     _      h 

h  +  y      3a  +  x 

ax  -{-  2hy  =  d. 
hcx  =  cy  —  2h. 


he 


Ans.    -j 


l         a 


a  +  h 
ac 


Ans. 


y 


+  h 

_  eg  —  hh 
ag-hf 

_ah—  ef 


Ans. 


J 


ag-bf 

_2h'-6a'-}-d 
3a 

_Sa'-h'  +  d 
3b 


be  c 


\ 


Ans. 


a 


a.  +  2h 


EQUATIONS   OF    THE    FIRST    DEGREE. 


135 


17. 


3.  +  5,  =  («|e^.l 


y  —  x- 


-2bf 
b'-f 


Ans. 


(  ¥ 


b+f 


Problems  for  Solution. 

1.    What  fraction  is  that  to  the  numerator  of  which  if  1  be 

added  the  value  will  be  -,  but  if  1  be  added  to  its  denominator 

1  ^ 

the  value  will  be  -  ? 
4 


Let 

-  =  the  fraction. 

y 

Then 

^+1      \and       ^         1. 

y      ^         y+i    4 

Whence 

3  a;  +  3  =  ?/,  and  4  a;  =  y  +  1. 

Subtracting, 

a;  —  3  =  1,  and  a;  =  4. 

Hence 

12  +  3  =  .v. 

.-.  y  =  15. 

2.  A  market-woman  bought  a  certain  number  of  eggs  at 
2  for  a  penny,  and  as  many  others  at  3  for  a  penny ;  and, 
having  sold  them  altogether  at  the  rate  of  5  for  2  pence,  she 
found  that  she  had  lost  4  pence.  How  many  of  both  kinds 
did  she  buy  ? 

Let  2x  =  the  whole  number  of  eggs. 

Then  x  =  the  number  of  eggs  of  each  sort. 


Then  will 
and 


1 


X  =  the  cost  of  the  first  sort, 


-  a;  =  the  cost  of  the  second  sort. 


But  by  the  conditions  of  the  question 

4  a; 
5* 


5:2ar::2: 


136  ELEMENTARY    ALGEBRA. 

Hence  — -  will  denote  the  amount  for  which  the  eggs  were  sold. 
5 

But  by  the  conditions 

1,1         4a;      . 

-  X  -^  -X =  4. 

2        3  5 

.-.  15a;  +  10a;-24a:  =  120. 

.-.  X  =  120,  the  number  of  eggs  of  each  sort. 

3.  A  person  possessed  a  capital  of  $30,000,  for  which  he 
received  a  certain  interest ;  but  he  owed  the  sum  of  $20,000, 
for  which  he  paid  a  certain  annual  interest.  The  interest 
that  he  received  exceeded  that  which  he  paid  by  $800. 
Another  person  possessed  $35,000,  for  which  he  received 
interest  at  the  second  of  the  above  rates ;  but  he  owed  $24,000, 
for  which  he  paid  interest  at  the  first  of  the  above  rates. 
The  interest  that  he  received  annually  exceeded  that  which 
he  paid  by  $310.     Kequired  the  two  rates  of  interest. 

Let  X  denote  the  number  of  units  in  the  first  rate  of  interest,  and  y 
the  unit  in  the  second  rate.  Then  each  may  be  regarded  as  denoting 
the  interest  on  $100  for  1  year. 

To  obtain  the  interest  of  $30,000  at  the  first  rate,  denoted  by  x,  we 
form  the  proportion 

100  :  30,000  -.-.x:  55222^    or  300  a;  ; 
100 

and  for  the  interest  of  $20,000,  the  rate  being  y, 

100  :  20,000  :  :  2/ :  ?9999l^  or  200y. 
•^      100  /  -^ 

But  by  the  conditions  the  difference  between  these  two  amounts  is 
equal  to  $800. 

We  have,  then,  for  the  first  equation  of  the  problem, 

300  a; -200?/ =  800. 

By  expressing  algebraically  the  second  condition  of  the  problem, 
we  obtain  a  second  equation, 

3503/ -240x- 310. 


EQUATIONS    OF    THE    FIRST    DEGREE.  137 

Both  members  of  the  first  equation  being  divisible  by  100,  and  those 
of  the  second  by  10,  we  have 

3a;-2y  =  8,    SSy  -  24a;  =  31. 

To  eliminate  x,  multiply  the  first  equation  by  8,  and  then  add  the 

result  to  the  second.     There  results 

192/ =  95,  whence  y  =  5. 

Substituting  for  y,  in  the  first  equation,  this  value,  and  that  equation 

becomes 

3  a;  —  10  =  8,  whence  a;  =  6. 

Therefore  the  first  rate  is  6  per  cent ;  and  the  second,  5. 

Veeification. 

$30,000  at  6  per  cent  =  30,000  X  .06  =  $1800, 
$20,000  at  5  per  cent  =  20,000  X  .05  =  $1000  ; 
and  we  have  1800  -  1000  =  800. 

The  second  condition  can  be  verified  in  the  same  manner. 

4.  What  two  numbers  are  those  whose  difference  is  7,  and 
sum  33?  Ans.  13  and  20. 

5.  Divide  the  number  75  into  two  such  parts  that  three 
times  the  greater  may  exceed  seven  times  the  less  by  15. 

Ans.  54  and  21. 

6.  In  a  mixture  of  wine  and  cider,  }  of  the  whole,  plus  25 
gallons,  was  wine ;  and  J  part,  minus  5  gallons,  was  cider. 
How  many  gallons  were  there  of  each  ? 

Ans.  85  of  wine,  and  35  of  cider. 

7.  A  bill  of  £120  was  paid  in  guineas  and  moidores,  and 
the  number  of  pieces  used  of  both  sorts  was  just  100.  If  the 
guinea  be  estimated  at  21s.,  and  the  moidore  at  27s.,  how 
many  pieces  were  there  of  each  sort  ?  Ans.  50. 

8.  Two  travelers  set  out  at  the  same  time  from  London  and 
York,  whose  distance  apart  is  150  miles.  One  of  them  travels 
8  miles  a  day ;  and  the  other,  7.  In  what  time  will  they 
meet?  Ans.  In  10  davs. 


138  ELEMENTARY    ALGEBRA. 

9.  At  a  certain  election  375  persons  voted  for  two  candi- 
dates, and  the  candidate  chosen  had  a  majority  of  91.  How 
many  voted  for  each  ? 

A71S.  233  for  one,  and  142  for  the  other. 

10.  A  person  has  two  horses,  and  a  saddle  worth  £50. 
Now,  if  the  saddle  be  put  on  the  back  of  the  first  horse,  it 
makes  their  joint  value  double  that  of  the  second  horse ;  but 
if  it  be  put  on  the  back  of  the  second,  it  makes  their  joint 
value  triple  that  of  the  first.  What  is  the  value  of  each 
horse?  Arts.  One  £30,  and  the  other  £40. 

11.  The  hour  and  minute  hands  of  a  clock  are  exactly 
together  at  12  o'clock.     When  will  they  again  be  together? 

Ans.  Ih.  5^m. 

12.  A  man  and  his  wife  usually  drank  a  cask  of  beer  in 
12  days ;  but  v/hen  the  man  was  from  home,  it  lasted  the 
woman  30  days.  How  many  days  would  the  man  alone  be 
in  drinking  it?  Ans.  20  days. 

13.  If  32  pounds  of  sea-water  contain  1  pound  of  salt,  how 
much  fresh  water  must  be  added  to  these  32  pounds  in  order 
that  the  quantity  of  salt  contained  in  32  pounds  of  the  new 
mixture  shall  be  reduced  to  2  ounces,  or  i  of  a  pound  ? 

Ans.  224  lbs. 

14.  A  person  who  possessed  $100,000  placed  the  greater 
■part  of  it  out  at  5  per  cent  interest,  and  the  other  at  4  per 
cent.  The  interest  which  he  received  for  the  whole  amounted 
to  $4640.    Required  the  two  parts.  Ans.  $64,000  and  $36,000. 

15.  At  the  close  of  an  election  the  successful  candidate 
had  a  majority  of  1500  votes.  Had  a  fourth  of  the  votes  of 
the  unsuccessful  candidate  been  also  given  to  him,  he  would 
have  received  three  times  as  many  as  his  competitor,  wanting 
three  thousand  five  hundred.  How  many  votes  did  each 
receive?  Ans.  1st,  6500;  2d,  5000. 


EQUATIONS  OF  THE  FIRST  DEGREE.         139 

16.  A  gentleman  bought  a  gold  and  a  silver  watch,  and  a 
chain  worth  $25.  When  he  put  the  chain  on  the  gold  watch, 
it  and  the  chain  became  worth  three  and  a  half  times  more 
than  the  silver  watch ;  but  when  he  put  the  chain  on  the 
silver  watch,  they  became  worth  one  half  the  gold  watch  and 
$15  over.     What  w^as  the  value  of  each  watch  ? 

^715.  Gold  watch,  $80  ;  silver  watch,  $30. 

17.  There  is  a  certain  number  expressed  by  two  figures, 
which  figures  are  called  digits.  The  sum  of  the  digits  is  11, 
and  if  13  be  added  to  the  first  digit  the  sum  will  be  three 
times  the  second.     What  is  the  number?  Arts.  56. 

18.  From*  a  company  of  ladies  and  gentlemen,  15  ladies 
retire,  and  there  are  then  left  2  gentlemen  to  each  lady;  after 
which  45  gentlemen  depart,  w^hen  there  are  left  5  ladies  to 
each  gentleman.     How  many  were  there  of  each  at  first? 

Ans.  50  gentlemen,  40  ladies. 

19.  A  person  wishes  to  dispose  of  his  horse  by  lottery.  If 
he  sells  the  tickets  at  $2  each,  he  will  lose  $30  on  his  horse ; 
but  if  he  sells  them  at  $3  each,  he  will  receive  $30  more  than 
his  horse  cost  him.  What  is  the  value  of  the  horse,  and  the 
number  of  tickets?  Ans.  Horse,  $150;  No.  of  tickets,  60. 

20.  A  person  purchases  a  lot  of  wheat  at  $1,  and  a  lot  of 
rye  at  75  cents,  per  bushel,  the  whole  costing  him  $117.50. 
He  then  sells  \  of  his  wheat  and  \  of  his  rye  at  the  same  rate, 
and  realizes  $27.50.     How  much  did  he  buy  of  each  ? 

Ans.  80  bushels  of  wheat,  50  of  rye. 

21.  There  are  52  pieces  of  money  in  each  of  two  bags. 
A  takes  from  one,  and  B  from  the  other.  A  takes  twice  as 
much  as  B  left,  and  B  takes  seven  times  as  much  as  A  left. 
How  much  did  each  take?     Ans.  A,  48  pieces;  B,  28  pieces. 

22.  Two  persons,  A  and  B,  purchase  a  house  together, 
worth  $1200.     Says  A  to  B,  "  Give  me  two  thirds  of  your 


140  ELEMENTARY    ALGEBRA. 

money,  and  I  can  purchase  it  alone."  But  says  B  to  A,  "If 
you  will  give  me  three  fourths  of  your  money,  I  shall  be  able 
to  purchase  it  alone."     How  much  had  each? 

Ans.  A,  $800 ;  B,  $600. 

23.  A  grocer  finds  that  if  he  mixes  sherry  and  brandy  in 
the  proportion  of  2  to  1,  the  mixture  will  be  worth  78s.  per 
dozen ;  but  if  he  mixes  them  in  the  proportion  of  7  to  2,  he 
can  get  79s.  a  dozen.  What  is  the  price  of  each  liquor  per 
dozen  ?  Ans.  Sherry,  81s. ;  brandy,  72s. 

Equations  containing  Three  or  More  Unknown  Quantities. 

117.  Let  us  now  consider  equations  involving  three  or  more 
unknown  quantities. 

Take  the  group  of  simultaneous  equations, 

6^~6y  +  4z  =  15  (1) 

'   7a:  +  4y-32;=:19  (2) 

2x+    2/+ 6^-^46  (3) 

To  eliminate  z  by  means  of  the  first  two  equations,  multiply 
the  first  by  3,  and  the  second  by  4.  Then,  since  the  coefficients 
of  z  have  contrary  signs,  add  the  two  results  together.  This 
gives  a  new  equation, 

43:r-2?/  =  121  (4) 

Multiplying  the  second  equation  by  2  (a  factor  of  the  co- 
efficient of  z  in  the  third  equation),  and  adding  the  result  to 
the  third  equation,  we  have 

16:r  +  9y  =  84  (5) 

The  question  is  then  reduced  to  finding  the  values  of  x  and 
y,  which  will  satisfy  tha  new  equations  (4)  and  (5). 

Now,  if  the  first  be  multiplied  by  9,  the  second  by  2,  and 
the  results  added  together,  we  find 

419a:  =-1257,  whence  a:  =  3. 


■  EQUATIONS   OF    THE    FIRST    DEGREE.  141 

We  might,  by  means  of  Equations  (4)  and  (5),  determine  y 
in  the  same  way  that  we  have  determined  x ;  but  the  value 
of  y  may  be  determined  more  simply,  by  substituting  the 
value  of  X  in  Equation  (5).     Thus, 

48  +  9y  =  84.  .-.^  =  84^  =  4. 

In  the  same  manner  the  first  of  the  three  given  equations 
becomes,  by  substituting  the  values  of  x  and  y, 

15-24  +  42-15.  .-.  2-:~=6. 

4 

In  the  same  way  any  group  of  simultaneous  equations  may 
be  solved.     Hence  the  rule  :  — 

Combine  one  equation  of  the  group  with  each  of  the  others, 
and  eliminate  one  unknown  quantity :  there  will  result  a  new 
group  containing  one  equation  less  than  the  original  group. 

Combine  one  equation  of  this  new  group  with  each  of  the 
others,  and  eliminate  a  second  unknown  quantity :  there  will 
result  a  new  group  containing  two  equations  less  than  the  origi- 
nal group.     J 

Continue  the  operation  until  a  single  equation  is  found,  con- 
taining  but  one  unknown  quantity. 

Find  the  value  of  this  unhwivn  quantity  by  the  preceding 
rules ;  substitute  this  in  one  of  the  group  of  two  equations,  and 
find  the  value  of  a  second  unknown  quantity ;  substitute  these 
in  either  of  the  groups  of  three,  finding  a  third  unknown  quan- 
tity;  and  so  on  till  the  values  of  all  are  found. 

Notes.  —  1.  In  order  that  the  value  of  the  unknown  quantities  may 
be  determined,  there  must  be  just  as  many  independent  equations  of 
condition  as  there  are  unknown  quantities.  If  there  are  fewer  equa- 
tions than  unknown  quantities,  the  resulting  equation  will  contain  at 
least  two  unknown  quantities,  and  hence  their  values  cannot  be  found 
(§  110).  If  there  are  more  equations  than  unknown  quantities,  the 
conditions  may  be  contradictory,  and  the  equations  impossible. 


142  ELEMENTARY   ALGEBRA. 

2.  It  often  happens  that  each  of  the  proposed  equations  does  not 
contain  all  the  unknown  quantities.  In  this  case  the  elimination  may 
often  be  very  readily  performed. 

Take  the  four  equations  involving  four  unknown  quanti- 
ties, 

2a;-3y  +  22-13         (1)  4^  +  22=14         (3) 

4t^-2:r-30        (2)  5y  +  3w  =  32        (4) 

By  inspecting  these  equations,  we  see  that  the  elimination 
of  z  in  the  two  equations  (1)  and  (3)  will  give  an  equation 
involving  x  and  y ;  and  if  we  eliminate  w  in  Equations  (2) 
and  (4),  we  shall  obtain  a  second  equation  involving  x  and  y. 
These  last  tw^o  unknown  quantities  may  therefore  be  easily 
determined.  In  the  first  place,  the  elimination  of  z  from  (1) 
and  (3)  gives 

7y-2^=-l. 
That  of  u  from  (2)  and  (4)  gives 

20y  +  6a;  =  38. 
Multiplying  the  first  of  these  equations  by  3,  and  adding, 

41y  =  41. 
Whence  y  —  1. 

Substituting  this  value  in 

73/-2a;  =  l, 
we  find  x  =  3. 

Substituting  for  x  its  value  in  Equation  (2),  it  becomes 

42^-6  =  30. 
Whence  w  =  9. 

Substituting  for  y  its  value  in  Equation  (3),  there  results 

2  =  5. 


EQUATIONS    OF    THE    FIRST    DEGREE. 


143 


1.    Given    < 


Exercises. 


-X  -\--y  4-  -z=^  10, 
2    ^3"^      4 


>  to  find  X,  y,  and  z. 


>  to  find  X,  y,  and  z. 


Ans.  :r  =  8,  y  —  9,  z  —  12. 

■2x  +  4:y-Sz  =  22,^ 

Given    J^  ix —  2y  -\- 5z  =  18,  >-  to  find  x,  y,  and  z. 

,6^+7y-    0  =  63,) 

^ns.  a;  =  3,  y  =  7,  z  =  4. 

3.    Given    ^ -rr +  -y +  -z  =  15, 

^725.  o:  =  12,  y  =  20,  2  =  30. 
r^  +  y  +  0  =  29i,^ 
Given    ■<  ^  +  y  —  2;  =  18^,  >-  to  find  x,  y,  and  z. 
\.x-y-\-z-=\Zl,) 

Ans.  :r  =  16,  y  =  7f ,  2  =  5^. 
r3:r  +  5y  =  161,-) 
Given    -(  7:r  +  2^  =  209,  >■  to  find  x,  y,  and  z. 
(2y+    2;=    89,3 

^715.  a:  =  17,  y  =  22,  2  =  45. 

-  +  -  =  «, 
a;     y 


4. 


5. 


6.    Given    - 


X       z 

yy    2 


^  to  find  X,  y,  and  2:. 


Ans.  X  ■ 


2 


y  = 


a  +  <?  —  ^ 


^  4-  ^  —  a 


144  ELEMENTARY   ALGEBRA. 

Note.  —  In  this  example  we  should  not  proceed  to  clear  the  equa- 
tion of  fractions,  but  subtract  immediately  the  second  equation  from  the 
first,  and  then  add  the  third.     We  thus  find  the  value  of  y. 


Problems  for  Solution. 

.  1.  Divide  the  nunlber  90  into  four  such  parts  that  the 
first  increased  by  2,  the  second  diminished  by  2,  the  third 
multiplied  by  2,  and  the  fourth  divided  by  2,  shall  be  equal 
each  to  each. 

Note. — This  problem  may  be  easily  solved  by  introducing  a  new 
unknown  quantity. 

Let  X,  y,  2,  and  u  denote  the  required  parts,  and  designate  by  m  the 
several  equal  quantities  which  arise  from  the  conditions.  We  shall  then 
have 

a;  +  2  =  m,   v  —  2  =  m,   2  2  =  m,   -  =  m. 

From  which  we  find 

By  adding  the  equations, 

x-{-y  +  z-\-u  — 171  + m^ —  -f-2m  =  4Jm. 

Since,  by  the  conditions  of  the  problem,  the  first  member  is  equal  to 
90,  we  have 

4im  =  90,   or   5^  =  90. 
Hence  m  =  20. 

Having  the  value  of  ?m,  we  easily  find  the  other  values ;  viz., 

a;  =  18,  y  =  22,  2  =  10,   w  =  40. 

2.  There  are  three  ingots,  composed  of  different  metals, 
mixed  together.  A  pound  of  the  first  contains  7  ounces  of 
silver,  3  ounces  of  copper,  and  6  of  pewter.  A  pound  of  the 
second  contains  12  ounces  of  silver,  3  ounces  of  copper,  and  1 
of  pewter.  A  pound  of  the  third  contains  4  ounces  of  silver, 
7  ounces  of  copper,  and  5  of  pewter.     It  is  required  to  find 


EQUATIONS    OF    THE    P^IRST    DEGREE.  145 

how  much  it  will  take  of  each  of  the  three  ingots  to  form  a 

fourth,  which  shall  contain  in  a  pound,  8  ounces  of  silver,  3} 

of  copper,  and  41  of  pewter. 

Let  X,  y,  and  z  denote  the  number  of  ounces  which  it  is  necessary  to 

take  from  the  three  ingots  respectively,  in  order  to  form  a  pound  of  the 

required  ingot.     Since  there  are  7  ounces  of  silver  in  a  pound  (or  1(3 

ounces)  of  the  first  ingot,  it  follows  that  one  ounce  of  it  contains  ^^  of  an 

ounce  of  silver,  and  consequently  in  a  number  of  ounces  denoted  by  x 

there  is  —  ounces  of  silver.     In  the  same  manner  we  find  that  — ^ 
16  16 

and  —  denote  the  number  of  ounces  of  silver  taken  from  the  second  and 

16 
the  third ;  but,  from  the  enunciation,  one  pound  of  the  fourth  ingot  con- 
tains 8  ounces  of  silver.     We  have,  then,  for  the  first  equation, 
7£     12?/  ,  42^g. 
16        16       16 
or,  clearing  fractions,        *7x  +  12y  +  4z  ==  128. 
As  respects  the  copper,  we  should  find 

3a;  +  3y  +  72  =  60; 
and  with  reference  to  the  pewter, 

6a;  +  y  +  52  =  68. 
As  the  coefficients  of  y  in  these  three  equations  are  the  most  simple, 
it  is  convenient  to  eliminate  this  unknown  quantity  first. 

Multiplying  the  second  equation  by  4,  and  subtracting  the  first  from 
it,  member  from  member,  we  have 

5a; +  242  =  112. 
Multiplying  the  third  equation  by  3,  and  subtracting  the  second  from 
the  resulting  equation,  we  have 
K  15a; +  82  =144. 

"^  Multiplying  this  last  equation  by  3,  and  subtracting  the  preceding 
one,  we  obtain 

40  a;  =  320. 

Whence  a;  =  8. 

Substitute  this  value  for  x  in  the  equation 

15a; +  82  =  144, 

it  becomes  120  +  8  2  =  144. 

Whence  2  =  3. 

D.  N.  E.  A.  —  10. 


146  ELEMENTARY    ALGEBRA. 

Lastly,  the  two  values,  x  =  8,  z  =  3,  being  substituted  in  the  equation 

6x  +  2/  +  52  =  68, 

give  48  +  2/  +  15  =  68. 

Whence  2/  =  ^• 

Therefore,  in  order  to  form  a  pound  of  the  fourth  ingot,  we  must  take 
8  ounces  of  the  first,  5  ounces  of  the  second,  and  3  of  the  third. 

Verification.  If  there  be  7  ounces  of  silver  in  16  ounces  of  the 
first  ingot,  in  8  ounces  of  it  there  should  be  a  number  of  ounces  of  silver 
expressed  by 

7x8 
16  * 

T    TT  12x5       ,  4x3 

In  like  manner,  and  

16  16 

will  express  the  quantity  of  silver  contained  in  5  ounces  of  the  second 
ingot,  and  3  ounces  of  the  third. 

Now,  we  have 

7x812x54x3_128_3 
16  16  16         16 

Therefore  a  pound  of  the  fourth  ingot  contains  8  ounces  of  silver,  as 
required  by  the  enunciation.  The  same  conditions  may  be  verified  with 
respect  to  the  copper  and  the  pewter. 

3.  A's  age  is  double  B's,  and  B's  is  triple  C's,  and  the  sum 
of  all  their  ages  is  140.     What  is  the  age  of  each  ? 

Am.  A's,  84 ;  B's,  42 ;  and  C's,  14. 

4.  A  person  bought  a  chaise,  horse,  and  harness  for  £60. 
The  horse  came  to  twice  the  price  of  the  harness  ;  and  the 
chaise,  to  twice  the  cost  of  the  horse  and  harness.  What  did 
he  give  for  each  ? 

Ans.  £13  6s.  8d.  for  the  horse  ;  £6  13s.  4d.  for  the  har- 
ness ;  £40  for  the  chaise. 

6.  Divide  the  number  36  into  three  such  parts  that  -J-  of 
the  first,  i  of  the  second,  and  }  of  the  third,  may  be  all  equal 
to  each  other.  Ans.  8,  12,  and  16. 


EQUATIONS    OF    THE    FIRST    DEGREE.  147 

6.  If  A  and  B  together  can  do  a  piece  of  work  in  8  days, 
A  and  C  together  in  9  days,  and  B  and  C  in  10  days,  how 
many  days  would  it  take  each  to  perform  the  same  work 
alone?  Ans.  A,  14||-  days;  B,  17|f ;  C,  233V 

7.  Three  persons,  A,  B,  and  C,  begin  to  play  together, 
having  among  them  all  $600.  At  the  end  of  the  first  game 
A  has  won  one  half  of  B's  money,  which,  added  to  his  own, 
makes  double  the  amount  B  had  at  first.  In  the  second  game 
A  loses  and  B  wins  just  as  much  as  C  had  at  the  beginning, 
when  A  leaves  off  with  exactly  what  he  had  at  first.  How 
much  had  each  at  the  beginning  ? 

Ans.  A,  $300  ;  B,  $200;  C,  $100. 

8.  Three  persons.  A,  B,  and  C,  together  possess  $3640. 
If  B  gives  A  $400  of  his  money,  then  A  will  have  $320  more 
than  B  ;  but  if  B  takes  $140  of  C's  money,  then  B  and  C  will 
have  equal  sums.     How  much  has  each  ? 

Ans.  A,  $800 ;  B,  $1280  ;  C,  $1560. 

9.  Three  persons  have  a  bill  to  pay,  which  neither  alone 
is  able  to  discharge.  A  says  to  B,  "  Give  me  the  fourth  of 
your  money,  and  then  I  can  pay  the  bill."  B  says  to  C, 
"  Give  me  the  eighth  of  yours,  and  I  can  pay  it."  But  C  says 
to  A,  "  You  must  give  me  the  half  of  yours  before  I  can  pay 
it,  as  I  have  but  $8."  What  was  the  amount  of  their  bill,  and 
how  much  money  had  A  and  B  ? 

Ans.  Amount  of  the  bill,  $13  ;  A  had  $10,  and  B  $12. 

10.  A  person  possessed  a  certain  capital,  which  he  placed 
out  at  a  certain  interest.  Another  person,  who  possessed 
$10,000  more  than  the  first,  and  who  put  out  his  capital  1  per 
cent  more  advantageously,  had  an  annual  income  greater  by 
$800.  A  third  person,  who  possessed  $5000  more  than  the 
first,  putting  out  his  capital  2  per  cent  more  advantageously, 


148  ELEMENTARY    ALGEBRA. 

had  an  annual  income  greater  by  $1500.     Required  the  capi- 
tals of  the  three  persons,  and  the  rates  of  interest. 

Ans.  Capitals,  $30,000,  $40,000,  $45,000 ;  rates  of  inter- 
est, 4%,  5%,  6%. 

11.  A  widow  receives  an  estate  of  $15,000  from  her  de- 
ceased husband,  with  directions  to  divide  it  among  two  sons 
and  three  daughters  so  that  each  son  may  receive  twice  as 
much  as  each  daughter,  she  herself  to  receive  $1000  more 
than  all  the  children  together.  What  was  her  share,  and 
what  the  share  of  each  child  ? 

Ans.  The  widow's  share,  $8000  ;  each  son's,  $2000 ;  each 

daughter's,  $1000. 

12.  A  certain  sum  of  money  is  to  be  divided  between  three 
persons.  A,  B,  and  C.  A  is  to  receive  $3000  less  than  half  of 
it ;  B,  $1000  less  than  one  third  part ;  and  C,  $800  more  than 
the  fourth  part  of  the  whole.  What  is  the  sum  to  be  divided, 
and  what  does  each  receive  ? 

Ans.  Sum,  $38,400;   A  receives  $16,200;    B,  $11,800; 

C,  $10,400. 

13.  A  person  has  three  horses,  and  a  saddle  which  is  worth 
$220.  If  the  saddle  be  put  on  the  back  of  the  first  horse,  it 
will  make  his  value  equal  to  that  of  the  second  and  third  ; 
if  it  be  put  on  the  back  of  the  second,  it  will  make  his  value 
double  that  of  the  first  and  third ;  if  it  be  put  on  the  back  of 
the  third,  it  will  make  his  value  triple  that  of  the  first  and 
second.     What  is  the  value  of  each  horse  ? 

Ans.  1st,  $20;  2d,  $100;  3d,  $140. 

14.  The  crew  of  a  ship  consisted  of  her  complement  of 
sailors,  and  a  number  of  soldiers.  There  were  22  sailors  to 
every  three  guns,  and  10  over ;  also  the  whole  number  of 
hands  was  five  times  the  number  of  soldiers  and  guns  to- 
gether.    But  after  an  engagement,  in  which  the  slain  were 


INEQUALITIES.  149 

one  fourth  of  the  survivors,  there  wanted  5  men  to  make  13 
men  to  every  two  guns.  Eequired  the  number  of  guns,  sol- 
diers, and  sailors. 

Ans.  90  guns,  55  soldiers,  and  670  sailors. 

15.  Three  persons  have  $96,  which  they  wish  to  divide 
equally  between  them.  In  order  to  do  this,  A,  who  has  the 
most,  gives  to  B  and  C  as  much  as  they  have  already  ;  then 
B  divides  with  A  and  C  in  the  same  manner,  that  is,  by  giv- 
ing to  each  as  much  as  he  had  after  A  had  divided  with  them  ; 
C  then  makes  a  division  with  A  and  B ;  when  it  is  found  that 
they  all  have  equal  sums.     How  much  had  each  at  first? 

Ans,  1st,  $52;  2d,  $28;  3d,  $16. 

16.  Divide  the  number  a  into  three  such  parts  that  the 

first  shall  be  to  the  second  as  m  to  n,  and  the  second  to  the 

third  as  p  to  q. 

amp  anp  ana 

Ans.  rr  = ....,   ,   .,  ,  ,       ,  3/~  — 


77ip  +  np  +  nq'  ^      mp  ^np  -\-nq'  ^      mp -{-np-\-nq 

17.    Three  masons,  A,  B,  and  C,  are  to  build  a  wall.     A 

and  B  together  can  do  it  in  12  days  ;  B  and  C,  in  20  days ; 

and  A  and  C,  in  15  days.     In  what  time  can  each  do  it  alone, 

and  in  what  time  can  they  all  do  it  if  they  work  together  ? 

Ans.  A,  in  20  days ;  B,  in  30 ;  and  C,  in  60 ;  all,  in  10. 

Inequalities. 

117  a.  An  inequality  is  an  algebraic  expression  of  two  un- 
equal quantities,  connected  by  the  sign  of  inequality.  Thus, 
a  >  ^  is  an  inequality,  showing  that  a  is  greater  than  h. 

Of  two  negative  quantities,  that  one  is  the  greater  alge- 
hraically  which  has  the  fewer  units. 

The  part  on  the  left  of  the  sign  is  called  the  first  member, 
and  the  part  on  the  right  the  second  member,  of  the  inequality. 

Two  inequalities  Subsist  in  the  same  sense  when  the  greater 


150  ELEMENTARY    ALGEBRA. 

quantity  is  in  the  first  member  of  both  or  in  the  second  mem- 
ber of  both ;  they  subsist  in  a  contrary  sense  when  the  greater 
quantity  is  in  the  first  member  of  one  and  in  the  second 
member  of  the  other.     Thus,  the  inequalities 

35>30  and  18  >  10 
subsist  in  the  same  sense,  and  the  inequalities 

15>13   and    12  <  14 

subsist  in  a  contrary  sense. 

The  following  principles  enable  us  to  transform  inequali- 
ties :  — 

(1)  If  we  add  the  same  quantity  to^  or  subtract  the  same 
quantity  Jrom,^  both  members  of  an  inequality^  the  resulting 
inequality  will  subsist  iyi  the  same  sense. 

Thus,  if  we  add  5  to,  and  subtract  5  from,  both  members 
of  the  inequality 

4>2, 

we  have  9  >  7   and   —  1  >  —  3. 

This  principle  enables  us  to  transpose  a  term  from  one 
member  of  an  inequality  to  the  other  by  simply  changing  its 
sign.     Thus,  from  the  inequality 

^x  —  b>2x  +  a, 
we  find,  by  transposition, 

x^  a-\-b. 

(2)  If  two  m>embers  of  an  inequality  be  m^ultiplied  or  divided 
by  a  positive  quantity,  the  resulting  inequality  will  subsist  in 
the  same  sense. 

Thus,  if  we  multiply  or  divide  both  members  of  the  in- 
equality 

12  >  8 

by  +  4,  we  have       48  >  32   and   3  >  2. 


INEQUALITIES.  151 

This  principle  enables  us  to  clear  an  inequality  of  fractions. 
Thus,  if  we  multiply  both  members  of  the  inequality 

4        3       6 

by  12,  and  reduce,  we  have 

9x-Sb>2x-S4:, 

The  following  precautions  are  to  be  observed  in  treating 
inequalities :  — 

(3)  If  both  members  of  an  inequality  are  positive,  they  TYiay 
he  raised  to  like  powers  without  changing  the  sense  of  the 
inequality. 

Thus,  if  both  members  of  the  inequality 

12  >  5 

be  squared  or  cubed,  we  have 

144  >  25  and  1728  >  125. 

(4)  If  the  two  members  of  an  inequality  be  multiplied  or 
divided  by  a  negative  quantity^  the  sense  of  the  inequality  must 
he  reversed. 

Thus,  if  we  multiply  and  divide  both  members  of  the  in- 
equality 

3<6 
by  —  3,  w^e  have 

-9>-18   and   -l>-2. 

(5)  If  both  members  of  an  inequality  are  negative,  they  may 
be  raised  to  any  power  of  an  odd  degree,  and  the  resulting 
inequality  will  subsist  in  the  same  sense ;  but,  if  both  "members 
be  raised  to  a  power  of  an  even  degree,  the  sense  of  the  resulting 
inequality  will  he  reversed. 


152  ELEMENTARY    ALGEBRA. 

Thus,  if  we  square  and  cube  both  members  of  the  inequality 

~3>-5, 

we  have  9<25   and   -27>-125. 

By  the  aid  of  these  principles  we  may  solve  an  inequality ; 
that  is,  we  can  find  an  inequality  in  wliich  the  unknown 
quantity  shall  form  one  member,  and  a  known  quantity  the 
other. 

Exercises. 

1.  5a:- 6  >  19.  Ans.  x>5. 

2.  3:r  +  -— :r-30>10.  Ans.  x>i. 

A 

6      3     ^2       2        2 

4. Y'bx  —  ah'>—'  Atis.  x^a. 

5  5 

6.    -~  —  ax-\-ah<—'  Ans.x<h. 

7  7 


CHAPTER  VI. 

PO"WERS. 

118.  A  power  of  a  quantity  is  the  product  obtained  by  tak- 
ing that  quantity  any  number  of  times  as  a  factor. 

If  the  quantity  be  taken  once  as  a  factor,  we  have  the  first 
power ;  if  taken  twice,  we  have  the  second  power ;  if  three 
times,  the  third  power ;  if  n  times,  the  nth.  power,  n  being 
any  whole  number  whatever. 

A  power  is  indicated  by  means  of  the  exponential  sign. 
Thus, 

a  =  a}  denotes  first  power  of  a. 

square,  or  second  power,  of  a. 
cube,  or  third  power,  of  a, 
fourth  power  of  a. 
fifth  power  of  a. 
mth  power  of  a. 


aX  a^  cv 


aX  aX  a 


_^3 


aXaXaXa  =  a*' 
aXaXaXaXa  =  a^ 
aX  aX  aX  a =  a"* 


Note.  —  Since  a^  =1  (?  49),  a^  x  a  =  1  x  a  =  a^  so  that  the  two 
factors  of  a^  are  1  and  a. 

In  every  power  there  are  three  things  to  be  considered :  — 

(1)  The  quantity  which  enters  as  a  factor,  and  which  is 
called  the  first  power. 

(2)  The  small  figure  which  is  placed  at  the  right  of  and 
a  little  above  the  letter,  which  is  called  the  exponent  of  the 
power,  and  shows  how  many  times  the  letter  enters  as  a 
factor. 

(3)  The  power  itself,  which  is  the  final  product,  or  result 
of  the  multiplications. 

153 


154  KLKMENTAIIV    ALGEBRA. 


Powers  of  Monomials. 

119.  Let  it  be  required  to  raise  the  monomial  2a^5^  to  the 
fourth  power.     We  have 

{2a'¥y  =  2a^h''  x  2a'h''  X  2a^h''  x  2a^b\ 

which  merely  expresses  that  the  fourth  power  is  equal  to  the 
product  which  arises  from  taking  the  quantity  four  times  as  a 
factor.     By  the  rules  for  multiplication,  this  product  is 

from  which  we  see 

(1)  That  the  coefficient  2  must  be  raised  to  the  fourth 
power ;  and 

(2)  That  the  exponent  of  each  letter  must  be  multiplied  by 
4,  the  exponent  of  the  power. 

As  the  same  reasoning  applies  to  every  example,  we  have, 
for  the  raising  of  monomials  to  any  power,  the  following 
rule : — 

liaise  the  coefficient  to  the  required  power. 
Multiply  the  exponent  of  each  letter  by  the  exponent  of  the 
power. 

(1)  What  is  the  square  of  3ay?  Ans.  9a*y^. 

(2)  What  is  the  cube  of  GaYx?  Ans.  2l6a'yx\ 

(3)  What  is  the  fourth  power  of  2ayi^^  Ans.   IGa'Y'b'''. 

(4)  What  is  the  square  of  a^b^  ^  Ans.  a*b^Y' 


POWER.S.  155 

(6)  What  is  the  seventh  power  of  o?bc(P  ?       Ans.  a^^b'^c'd^^. 

(6)  What  is  the  sixth  power  of  a^b^^df         Ans.  a''b'^c''d\ 

(7)  What  are  the  square  and  the  cube  of  —  2a'^b^? 

Square.  Cube. 

~2a'b''  -2a'b' 

-2a'b'  -2a'b' 


-2a'b' 


Sa'b' 


By  observing  the  way  in  which  the  powers  are  formed,  we 
may  conclude  that 

When  the  monomial  is  positive,  all  the  powers  will  be  positive. 
When  the  monomial  is  negative,  all  even  powers  will  be 
positive,  and  all  odd  powers  will  be  negative. 

Exercises. 

1 .  What  is  the  square  of  —  2  a*b^  ?  Ans.  4  a^b^^. 

2.  What  is  the  cube  of  -  ba^'b'?  Ans.  -  125a'"5«. 

3.  What  is  the  eighth  power  of  —  a^xi/^?        Ans.  +  a'^Vy". 

4.  What  is  the  seventh  power  of  —  a'^b^'c  ?    Ans.  —  a}'^V''c\ 

5.  What  is  the  sixth  power  of  2abY  ?  Ans.  64:a^b^y^. 

6.  What  is  the  ninth  power  of  —  a'^bc^?  Ans.  —  a^b^c^^. 

7.  What  is  the  sixth  power  of  -  3ab'dr       Ans.  729  a^b'^d\ 

8.  What  is  the  square  of  -  10a"»5V?  Ans.  100 a'^'b^c^. 

9.  What  is  the  cube  of  -  da'^b'^dT^    Ans.  ~  729 a'"»i^c£*r. 


156  ELEMENTARY    ALGEBRA. 

10.  What  is  the  fourth  power  of  —  4a^6W^  ? 

1 1 .  What  is  the  cube  of  -  4  a^'^b'^'c'd  ?     Ans.  -  64  a^H^c^d\ 

12.  What  is  the  fifth  power  of  2oJ'h''xy  f        Ans.  S2a''b'Vi/. 

13.  What  is  the  square  of  20a;"yV?  Ans.  400 :r'y *"6''^ 

14.  What  is  the  fourth  power  of  3  a"^'V  ?      Ans.  81  a*"5'"c'l 

15.  What  is  the  fifth  power  of  —  c^d^'^xY  ? 

Ans.  —  c'''d^^"'x'Y\ 

16.  What  is  the  sixth  power  of  -  arh'^^'c'^  ?        Ans.  a^^>''"c^. 

17.  What  is  the  fourth  power  of- 2  aVc?'?     Ans.  16aVc?'\ 

Powers  of  Fractions. 

120,  From  the  definition  of  a  power,  and  the  rule  for  the 
multiplication  of  fractions,  the  cube  of  the  fraction  -  is  written 

and  since  any  fraction  raised  to  any  power  may  be  written 
under  the  same  form,  we  find  any  power  of  a  fraction  by  the 
following  rule :  — 

Raise  the  numerator  to  the  required  power,  for  a  new 
numerator;  o/nd  the  denominator  to  the  required  power ^  for 
a  new  denominator. 

The  rule  for  signs  is  the  same  as  in  the  last  section. 


POWERS.  157 


Kxercises. 
Find  the  powers  of  the  following  fractions :  — 

,     fa-c\  J        o'-2ac  +  c' 

^-    [rTc)  '  b^+2bc  +  c' 

'•  KuJ-j  ■  ^"^-  ^• 

6.  /^y.  ^ns.  ^^'. 

/_3ay^Y  .Slay- 

8.  fourth  power  of  ^-^.  ^«^-  16^- 

9.  Cube  of  ^^.  ^ns.  ^~!?  + f  ^■^'"■^ 

o;  +  3/  ^'^  +  3  ^^y  +  3  X2/^  -j-  y' 

10.  Fourth  power  of Ans.  — — - — -• 

^                 ^a^if  16  a  V^ 

11.  tiftn  power  of  — -— Ans.  — ^^   ,   .  - 


158  ELEMENTARY    ALGEBRA. 


Powers  of  Binomials. 

121.    A  binomial,  like  a  monomial,  may  be  raised  to  any 
power  by  the  process  of  continued  multiplication. 

(1)    Find  the  fifth  power  of  the  binomial  a  +  6. 

a  +  6 1st  power. 

a  +  b 


a^  +    ab 
+    ab  +  b'^ 

a^-\-2ab  +b'^ 2d  power. 

a  +  6 


a^  +  2  a^b  +  ab'^ 

+    a^b-r  2ab''  +6^ 

a-+3a26+  3ab^  +b''^    .     .     .     .     3d  power. 
a  +b 


■3a^b+    3aW-\-      a¥ 

■    a^b-\-    Sa'b^+    Sab^  +  b* 


a*  +  4a36  +  (ya^b'^  +  4a5^  +  M       4th  power. 
a  +b 

a^  +  ia*b+  6a^b'^+  ^d'b^  +    ab* 

+    a^b  +  AaW  +  C^a'^b^  +  4a6*  -f  b^ 


a*  +  5  a*b  +  10  a^b^  +  10  a^b^  +  5  a6*  +  b^    Ans, 

122.  Note.  —  It  will  be  observed  that  the  number  of  multiplications 
is  always  1  less  than  the  number  of  units  in  the  exponent  of  the  power. 
Thus,  if  the  exponent  is  1,  no  multiplication  is  necessary.  If  it  is  2,  we 
multiply  once ;  if  it  is  3,  twice  ;  if  4,  three  times  ;  etc.  The  powers  of 
polynomials  may  be  expressed  by  means  of  an  exponent.  Thus,  to  ex- 
press that  a  +  Z>  is  to  be  raised  to  the  fifth  power,  we  write  (a  +  bf  ;  if 
to  the  mth  power,  we  write  (a  +  6)"*. 


POWERS.  169 


(2)  Find  the  fifth  power  of  the  binomial  a—  b. 

a  —b 1st  power. 

a  —  b 


ab 

ab  +¥ 


•2a6  +6^ 2d  power. 

■b 


-2a'^b+  ab'' 

-    a^b  +  2  a})'  -  P 

-Za^b  -v  2>ah'  —W   .     .     .     .      3d  power. 
■b 


-  a^b+    Za^b""-    3ab^  +  M'   - 

a*  —  4  a^6  +    6  a^i^  —   4  a6^  +  6*      4lh  power. 
a  —b 

a5_4a*6+    Qd^b''-    4:a'^b^  +    ab* 

-  a*b+    4:  a'^b''  -    6  a'^b^  +  ^a¥  -  b^ 

a^-5  a^b  +  10  d^b^  -  10  a'^b^  -f  5  a6*  -  b^    Ans. 

In  the  same  way  the  higher  powers  may  be  obtained.  By 
examining  the  powers  of  these  binomials,  it  is  plain  that  four 
things  must  be  considered,  — 

I.  The  number  of  terms  of  the  power. 

II.  The  signs  of  the  terms. 

III.  The  exponents  of  the  letters. 

IV.  The  coefficients  of  the  terms. 

Let  us  see  according  to  what  laws  these  are  formed. 

I.  The  Number  of  Terms. 

123.  By  examining  the  several  multiplications,  we  shall 
observe  that  the  first  power  of  a  binomial  contains  two  terms ; 
the  second  power,  three  terms ;  the  third  power,  four  terms ; 


160  ELEMENTARY    ALGEBRA. 

the  fourth  power,  five  ;  the  fifth  power,  six ;  etc.     Hence  we 
may  conclude  that 

The  number  of  terms  in  any  power  of  a  binomial  is  greater 
by  one  than  the  exponent  of  the  power. 

II.  The  Signs  of  the  Terms. 

124.  It  is  evident  that  when  both  terms  of  the  given  bino- 
mial are  plus,  all  the  terms  of  the  power  will  be  plus. 

If  the  second  term  of  the  binomial  is  negative,  then  all 
the  odd  terms,  counted  from  the  left^  will  be  positive^  and  all 
the  even  terms  negative. 

III.  The  Exponents. 

125.  The  letter  which  occupies  the  first  place  in  a  binomial 
is  called  the  leading  letter.  Thus,  a  is  the  leading  letter  in 
the  binomials  a  +  b  and  a  —  b. 

(1)  It  is  evident  tjiat  the  exponent  of  the  leading  letter  in 
the  first  term  will  be  the  same  as  the  exponent  of  the  power, 
and  that  this  exponent  will  diminish  by  one  in  each  term  to 
the  right  until  we  reach  the  last  term,  when  it  will  be  0 
(§  49). 

(2)  The  exponent  of  the  second  letter  is  0  in  the  first  term, 
and  increases  by  one  in  each  term  to  the  right,  to  the  last 
term,  when  the  exponent  is  the  same  as  that  of  the  given 
power. 

(3)  The  sum  of  the  exponents  of  the  two  letters  in  any 
term  is  equal  to  the  exponent  of  the  given  power.  This  last 
remark  will  enable  us  to  verify  any  result  obtained  by  means 
of  the  binomial  formula. 

Let  us  now  apply  these  principles  in  the  two  following 
examples,  in  which  the  coefficients  are  omitted  :  — 

{a  +  by  ,  ,  .  a'  +  a'b  +  a'b'  +  a'b'  +  a'b'  +  ab'  +  5«, 
(a-by  .  .  .  a'  -  a'b  +  a'b'  -  a'b'  +  a'b'  -  ab'  +  b\ 


POWERS.  161 

As  the  pupil  should  be  practiced  in  writing  the  terms  with 
their  proper  signs  without  the  coefficients,  we  will  add  a  few 


more  examples 

1.  {a  +  bf  . 

2.  (a  -  by  . 

3.  (a  +  bf  . 

4.  (a -by  . 


in  which  the  coefficients  are  not  given. 
.  a'  +  a'b  +  ab'  +  b\ 
.  a'-a'b  +  a'b'-ab'  +  b\ 
.  a'  +  a'b  +  a'b'  +  a'b'  +  ab'  +  b'. 
.  a'  -  a'b  +  a'b'  -  a'b'  +  aW  -  a'b'  +  ab'  -  b\ 


IV.   The  Coefficients. 

126.  The  coefficient  of  the  first  term  is  1.  The  coefficient 
of  the  second  term  is  the  same  as  the  exponent  of  the  given 
power.  The  coefficient  of  the  third  term  is  found  by  multi- 
plying the  coefficient  of  the  second  term  by  the  exponent  of 
the  leading  letter  in  that  term,  and  dividing  the  product 
by  2.     Finally, 

If  the  coefficient  of  any  term  be  *niulti'plied  by  the  exponent 
of  the  leading  letter  in  that  term,  and  the  product  divided  by 
the  number  which  marks  the  place  of  the  term  from  the  left^ 
the  quotient  will  be  the  coefficient  of  the  next  term. 

Thus,  to  find  the  coefficients  in  the  example 
{a-by  ,  ,  ,  a^  -Vb  -ji^^5^  -  a'b^  +  a^b'  -  a'b'  +  ab^  -  b\ 
we  first  place  the  exponent  7  as  a  coefficient  of  the  second 
term.     Then,  to  find  the   coefficient  of  the  third  term,  we 
multiply  7  by  6,  the  exponent  of  a,  and  divide  by  2.     The 
quotient,  21,  is  the  coefficient  of  the  third  term.     To  find  the 
coefficient  of  the  fourth,  we  multiply  21  by  6,  and  divide 
the  product  by  3  :  this  gives  35.     To  find  the  coefficient  of 
the  fifth  term,  we  multiply  35  by  4,  and  divide  the  product 
by  4  :  this  gives  35.     The  coefficient  of  the  sixth  term,  found 
in  the  same  way,  is  21 ;  that  of  the  seventh,  7  ;  and  that  of 
the  eighth,  1.     Collecting  these  coefficients, 
(a  -  by  =  a'~  7  a'b  + 21  a'b'-  35  a'b'+S5  a'b*-21  a'b'+7  ab'-b\ 

D.  N.  E   A.  —  11. 


162  ELEMENTARY    ALGEBRA. 

Note.  —  We  see,  in  examining  this  last  result,  that  the  coefficients  of 
the  extreme  terms  are  each  1,  and  that  the  coefficients  of  terms  equally  dis- 
tant from  the  extreme  terms  are  equal.  It  will  therefore  be  sufficient  to 
find  the  coefficients  of  the  first  half  of  the  terms,  and  from  these  the 
others  may  be  immediately  written. 

Exercises. 

1.  Find  the  fourth  power  of  a-\-h, 

Ans.  a*  +  4  a'Z)  +  6  a'b''  +  4  a5'  +  h'. 

2.  Find  the  fourth  power  of  a  —  b. 

Ans.  a*  ~  4 a'^  +  6 a'b''  -^ab^  +  b\ 

3.  Find  the  fifth  power  of  a  +  b. 

Ans.  a^  +  5  a^^  +  10  a^b''  +  10  a'b^  +  bab'  +  b\ 

4.  Find  the  fifth  power  of  a  —  b. 

Ans.  a^  -  5  a'b  +  10  d'b''  -  10  a^b^  ■\-hab'-  b\ 

5.  Find  the  sixth  power  of  a-\-b. 

Ans.  a«  +  6  d'b  +  15  a'b''  +  20  a^b^  +  15  a'Z>*  +  6  aZ>^  +  b\ 

6.  Find  the  sixth  power  of  a  —  b. 

Ans.  a""-^  a'^b  +  15  a'b''  -  20  a^b^  +  15  a^b"  -  6  a5^  +  b\ 

127.  When  the  terms  of  the  binomial  have  coefficients,  we 
may  still  write  out  any  power  of  it  by  means  of  the  binomial 
formula. 

(1)   Let  it  be  required  to  find  the  cube  of  2c  +  3c?. 

(a  +  Vf  =  a^  +  3a2&  +  Sai^  +  b\ 

Here  2  c  takes  the  place  of  a  in  the  formula,  and  3  d  the  place  of  b. 
Hence  we  have 

(2c  +  3Jf  =  (2c)3  +  3.(2c)2.3(^  +  3(2c)(3^)2  +  (3(/)^  (1) 

and  by  performing  the  indicated  operations  we  have 

(2c  +  3 df  =  8 c3  +  3Gc2J  -f  51  ci^2  ^  27 d^  (2) 


POWERS.  163 

If  we  examine  the  second  member  of  Equation  (1),  we  see 
that  each  term  is  made  up  of  three  factors,  —  first,  the  numer- 
ical factor;  second,  some  power  of  2c;  and,  third,  some 
power  of  3c?.  The  powers  of  2c  are  arranged  in  descending 
order  towards  the  right,  the  last  term  involving  the  0  power 
of  2  c,  or  1 ;  the  powers  of  3  c?  are  arranged  in  ascending  order 
from  the  first  term,  where  the  0  power  enters,  to  the  last  term. 

The  operation  of  raising  a  binomial  involving  coeflicients 
is  most  readily  eflfected  by  writing  the  three  factors  of  each 
term  in  a  vertical  column,  and  then  performing  the  multipli- 
cations as  indicated  below. 

Find  by  this  method  the  cube  of  2c  +  3c?. 


1+3       +3       +1 

Coefficients. 

8c' +    4c'    +    2c     +    1 

Powers  of  2  c. 

1     +    Sd    +    9d'  +21  d^ 

Powers  of  3  d. 

(2c  +  df  =  8c'  +  36cW+  54cc^'  +  21  d^ 

The  preceding  operation  hardly  requires  explanation.  In 
the  first  line  write  the  numerical  coefficients  corresponding  to 
the  particular  power,  in  the  second  line  write  the  descending 
powers  of  the  leading  term  to  the  0  power,  and  in  the  third 
line  write  the  ascending  powers  of  the  following  term  from 
the  0  power  upwards.  It  will  be  easiest  to  commence  the 
second  line  on  the  right  hand.  The  multiplication  should  be 
performed  from  above  downwards. 

(2)    Find  the  fourth  power  of  3  aV  —  2  hd. 

{a  +  hf  =  a*  +  4a36  4-  Ga'^J^  ^^a¥  +  h\ 

1+4  +6.  +4  +1 

81aV+    27  aV      +      9a*c2        +    Sa^c        +1 
1        ~     2bd       +      462^^2       _    S¥d^       +16 /A/ 

81  a^c^  -  216  a^c'bd  +  216  a*c''b''d^  -  96  a'cb^d^  +  16  //i* 


164  ELEMENTARY    ALGEBRA. 


Exercises. 


1.  What  is  the  cube  of  3a;  -  6y  ; 

Ans.  27 re'  -  162:rV  +  S2Axi/  -  216y\ 

2.  What  is  the  fourth  power  of  a  —  ?>hf 

Ans.  a'  -  I2a'b  +  b4:a'b'  -  lOS ab'  +  81  b\ 

3.  What  is  the  fifth  power  of  c  —  2d^ 

Ans.  c'  -  10  c*d  +  40  c'd'  -  80  c'd'  +  80  cd*  -  32  d\ 

4.  What  is  the  cube  of  5a  — Sd? 

Ans,  125  a'  -  225  a^c^  +  135  ad'  -  27  cf . 


CHAPTER  VII. 

EXTRACTION   OF  ROOTS,   AND   RADICALS. 

Extraction  of  Roots. 

128.  Evolution  is  the  process  of  finding  the  equal  factors  of 
a  number  or  quantity.  One  of  such  factors  is  called  a  root 
of  the  quantity,  and  the  operation  of  finding  it  is  called 
extracting  the  root. 

The  sign  V  is  called  the  radical  sign.  A  number  placed 
above  and  in  the  opening  of  the  radical  sign  indicates  the 
root  to  be  extracted,  and  is  called  the  index  of  the  root.  Thus, 
in  VSl,  3  is  the  index  of  the  root,  and  shows  that  the  cube 
root  is  to  be  found  ;  in  -\/'^2a}%^,  5  is  the  index  of  the  root. 
When  the  index  is  2,  it  may  be  omitted. 

Instead  of  the  radical  sign  and  index  to  denote  a  root,  a 
fractional  exponent  is  often  used,  in  which  the  index  is  the 
denominator.  Thus,  instead  of  Va,  we  may  use  or  to  denote 
the  cube  root  of  a;  also  Va^  =  a^,  '\fh^  =  ^*,  etc. 

The  square  root  of  a  number  is  one  of  its  two  equal  factors. 
Thus,  6  X  6  =  36  :  therefore  6  is  the  square  root  of  36. 

The  symbol  for  the  square  root  is  V  or  the  fractional 
exponent  ^.     Thus, 

Va,  or  a^, 
indicates  the  square  root  of  a,  or  that  one  of  the  two  equal 
factors  of  a  is  to  be  found. 

Square  Boot. 

129.  A  perfect  square  is  any  number  which  can  be  resolved 
into  iiuo  equal  integral  factors. 

165 


IGG  ELEMENTARY    ALGEBRA. 

The  following  table,  verified  by  actual  multiplication,  in- 
dicates all  the  perfect  squares  between  1  and  100  :  — 

Squares  ...  1,   4,    9,    16,   25,   36,   49,    64,    81,    100. 
Boots  ....  1,   2,   3,    4,     5,     6,     7,     8,     9,     10. 

We  may  employ  this  table  for  finding  the  square  root  of 
any  perfect  square  between  1  and  100. 

Look  for  the  number  in  the  first  line.  If  it  is  found  there, 
its  square  root  will  be  found  immediately  under  it. 

If  the  given  number  is  less  than  100,  and  not  a  perfect 
square,  it  will  fall  between  two  numbers  of  the  upper  line,  and 
its  square  root  will  be  found  between  the  two  numbers  directly 
below.  The  lesser  of  the  two  will  be  the  entire  part  of  the 
root,  and  will  be  the  true  root  to  within  less  than  1. 

Thus,  if  the  given  number  is  55,  it  is  found  between  the 
perfect  squares  49  and  64,  and  its  root  is  7  and  a  decimal 
fraction. 

Note.  —  There  are  ten  perfect  squares  between  1  and  100,  if  we  in- 
clude both  numbers  ;  and  eight,  if  we  exclude  both. 

If  a  number  is  greater  than  100,  its  square  root  will  be 
greater  than  10 ;  that  is,  it  will  contain  tens  and  units.  Let 
N  denote  such  a  number,  x  the  tens  of  its  square  root,  and 
y  the  units.     Then  will 

N={x  +  yY  =  x'  +  2xy  +  f  =  x'  +  {2x  +  y)y; 
that  is,  the  number  is  equal  to  the  square  of  the  tens  in  its 
roots,  plus  twice  the  product  of  the  tens   by   the  units,  plus 
the  square  of  the  units. 

For  example,  extract  the  square  root  of  6084. 

Since  this  number  is  composed  of  more   than 
two  places  of  figures,  its  root  will  contain  more       60  84 
than  one  :  but  since  it  is  less  than  10,000,  which 
is  the  square  of  100,  the  root  will  contain  but  two  figures ; 
that  is,  units  and  tens. 


ROOTS.  167 

Now,  the  square  of  the  tens  must  be  found  in  the  two  left- 
hand  figures,  which  we  will  separate  from  the  other  two 
by  putting  a  point  over  the  place  of  units,  and  a  second  over 
the  place  of  hundreds.  These  parts,  of  two  figures  each, 
are  called  periods.  The  part  60  is  comprised  between  the 
two  squares  49  and  64,  of  w^hich  the  roots  are  7  and  8  :  hence 
7  expresses  the  number  of  tens  sought;  and  the  required  root 
is  composed  of  7  tens  and  a  certain  number  of  units. 

The  figure  7  being   found,  we 
write  it  on  the  right  of  the  given  60  84  78 

number,  from  which  we  separate  49 

it  by  a  vertical   line:    then   we  7  X  2  =  14  8|  118  4 

subtract  its  square,  49,  from  60,  118  4 


which  leaves  a  remainder  of  11,  0 

to  which  we  bring  down  the  two 

next  figures,  84.  The  result  of  this  operation,  1184,  contains 
tivice  the  product  of  the  tens  by  the  units ^  plus  the  square  of 
the  units. 

But  since  tens  multiplied  by  units  cannot  give  a  product 
of  a  less  unit  than  tens,  it  follows  that  the  last  figure,  4,  can 
form  no  part  of  the  double  product  of  the  tens  by  the  units. 
This  double  product  is  therefore  found  in  the  part  118,  which 
we  separate  from  the  units'  place,  4. 

Now,  if  we  double  the  tens,  which  gives  14,  and  then 
divide  118  by  14,  the  quotient  8  will  express  the  units,  or  a 
number  greater  than  the  units.  This  quotient  can  never  be 
too  small,  since  the  part  118  will  be  at  least  equal  to  twice 
the  product  of  the  tens  by  the  units  ;  but  it  may  be  too 
large,  for  the  118,  besides  the  double  product  of  the  tens  by 
the  units,  may  likewise  contain  tens  arising  from  the  square 
of  the  units.  To  ascertain  if  the  quotient  8  expresses  the  right 
number  of  units,  we  write  the  8  on  the  right  of  the  14,  which 
gives  148,  and  then  we  multiply  148  by  8.  This  multiplica- 
tion, being  effected,  gives  for  a  product  1184,  a  number  equal 


168  ELEMENTARY    ALGEBRA. 

to  the  result  of  the  first  operation.  Having  subtracted  the 
product,  we  find  the  remainder  equal  to  0  :  hence  78  is  the 
root  required.  In  this  operation  we  form,  first,  the  square  of 
the  tens  ;  second,  the  double  product  of  the  tens  by  the  units  ; 
and,  third,  the  square  of  the  units.  » 

Indeed,  in  the  operations,  we  have  merely  subtracted  from 
the  given  number  6084 :  first,  the  square  of  7  tens,  or  of  70 ; 
second,  twice  the  product  of  70  by  8 ;  and,  third,  the  square 
of  8,  that  is,  the  three  parts  which  enter  into  the  composition 
of  the  square,  70  +  8,  or  78 ;  and  since  the  result  of  the  sub- 
traction is  0,  it  follows  that  78  is  the  square  root  of  6084. 

130.  The  operations  in  the  last  example  have  been  per- 
formed on  but  two  periods;  but  it  is  plain  that  the  same 
methods  of  reasoning  are  equally  applicable  to  larger  num- 
bers, for,  by  changing  the  order  of  the  units,  we  do  not 
change  the  relation  in  which  they  stand  to  each  other. 

Thus,  in  the  number  60  84  95,  the  two  periods  60  84  have 
the  same  relation  to  each  other  as  in  the  number  60  84 ;  and 
hence  the  methods  used  in  the  last  example  are  equally 
applicable  to  larger  numbers. 

131.  Hence,  for  the  extraction  of  the  square  root  of  num- 
bers, we  have  the  following  rule  :  — 

Point  off  the  given  number  into  periods  of  two  figures  each, 
beginning  at  the  right  hand. 

Note  the  greatest  perfect  square  in  the  first  period  on  the 
left,  and  place  its  root  on  the  right,  after  the  manner  of  a 
quotient  in  division  ;  then  subtract  the  square  of  this  root  froni^ 
the  first  period,  and  bring  down  the  second  period  for  a  re- 
mainder. 

Double  the  root  already  found,  and  place  the  result  on  the 
left  for  a  divisor.  Seek  how  Tnany  times  the  divisor  is  contained 
in  the  remainder,  exclusive  of  the  right-hand  figure,  and  place 
the  figure  in  the  root  and  also  at  the  right  of  the  divisor. 


ROOTS. 


169 


Multiply  the  divisor  thus  augmented  by  the  last  figure  of 
the  root,  and  subtract  the  product  from  the  remainder,  and 
bring  down  the  next  period  for  a  new  remainder.  But  if  any 
of  the  products  should  be  greater  than  the  remainder,  diminish 
the  last  fi.gure  of  the  root  by  one. 

Double  the  whole  root  already  found,  for  a  new  divisor, 
and  continue  the  operation  as  before  until  all  the  periods  are 
brought  down. 

132.  If,  after  all  the  periods  are  brought  down,  there  is  no 
remainder,  the  given  number  is  a  perfect  square. 

The  number  of  places  of  figures  in  the  root  will  always 
be  equal  to  the  number  of  periods  into  which  the  given  num- 
ber is  separated. 

If  the  given  number  has  not  an  exact  root,  there  will  be 
a  remainder  after  all  the  periods  are  brought  down  ;  in  which 
case  ciphers  may  be  annexed,  forming  new  periods,  for  each 
of  which  there  will  be  one  decimal  place  in  the  root. 


Exercises. 


1.    What  is  the  square  root  of  36729  ? 


Note.  —  In  this  example  two 
periods  of  decimals  were  annexed, 
and  hence  two  places  of  decimals  in 
the  root. 


2.  Find  the  square  root  of  7225. 

3.  Find  the  square  root  of  17689. 


29 

3  67  291191.64+ 

1 

267 

261 

381 

629 
381 

382  6 

24800 
22956 

3832  4 

184400 
153296 

31104 


Ans.  85. 
Ans.  133. 


170 


ELEMENTARY    ALGEBRA. 


4.  Find  the  square  root  of  994009. 

5.  Find  the  square  root  of  85673536. 

6.  Find  the  square  root  of  67798756. 

7.  Find  the  square  root  of  978121. 

8.  Find  the  square  root  of  956484. 

9.  What  is  the  square  root  of  36372961  ? 

10.  What  is  the  square  root  of  22071204? 

11.  What  is  the  square  root  of  106929  ? 

12.  What  of  12088868379025? 

13.  What  of  2268741  ? 

14.  What  of  7596796  ? 

15.  What  is  the  square  root  of  96  ? 

16.  What  is  the  square  root  of  153  ? 

17.  What  is  the  square  root  of  101  ? 

18.  What  of  285970396644  ? 

19.  What  of  41605800625  ? 

20.  What  of  48303584206084  ? 


Ans.  997. 

Ans.  9256. 

Am.  8234. 

Ans.  989. 

Ans.  978. 

A71S.  6031. 

Ans.  4698. 

Ans.  327. 

Ans.  3476905. 

Ans.  1506.23+. 

J^ns.' 2756.22+. 

Ans.  9.79795+. 

Ans.  12.36931+. 

Ans.  10.04987+. 

Ans.  534762. 

Ans.  203975. 

Ans.  6950078. 


133.  Since  the  square  or  second  power  of  a  fraction  is  ob- 
tained by  squaring  the  numerator  and  denominator  separately, 
it  follows  that 

The  square  root  of  a  fraction  will  he  equal  to  the  square  root 
of  the  numerator  divided  hy  the  square  root  of  the  denominator. 

For  example,  the  square  root  of  ^  is  equal  to  f  .•  for 

0  0 

?  V  -  —  —  ' 

b      b~  b'' 


ROOTS.  171 


Exercises. 


1.  What  is  the  square  root  of  -?  Ans.  — 

9  3 

2.  What  is  the  square  root  of  -— ?  Ans.  — 

16  4 

3.  What  is  the  square  root  of  —  ?  Ans.  — 

^81  9 

4.  What  is  the  square  root  of ?  Ans.  — 

^  361  19 

1 A  1 

•  5.    What  is  the  square  root  of  :— ?  Ans.  — 

^64  2 

6.  What  is  the  square  root  of ?  Ans. 

^  61009  247 

7.  What  is  the  square  root  of  — — — ?  Ans. 

^  956484  978 

134.  If  the  numerator  and  denominator  are  not  perfect 
squares,  the  root  of  the  fraction  cannot  be  exactly  found. 
We  can,  however,  easily  find  the  approximate  root  by  the 
following  rule :  — 

Multiply/  both  terms  of  the  fraction  hy  the  denominator ; 
then  extract  the  square  root  of  the  numerator,  and  divide  this 
root  hy  the  square  root  of  the  denominator.  The  quotient  will 
he  the  approximate  root. 


3 

1.    Find  the  square  root  of  —  Ans.  .7745+. 

5 


Multiplying  the  numerator  and  denominator  by  5, 
Hence  (3,8729+)  -^  5  =  .7745+. 


172 


ELEMENTARY    ALGEBRA. 


2.  What  is  the  square  root  of  -  ? 

14 

3.  What  is  the  square  root  of  — ? 

4.  What  is  the  square  root  of  Hy^? 

5.  What  is  the  square  root  of  7-^? 

6.  What  is  the  square  root  of  8^? 

5 

7.  What  is  the  square  root  of  --? 

8.  What  is  the  square  root  of  10y\? 


Ans.  1.32287+. 

Arts.  1.24721+. 

Ans.  3.41869+. 
Ans,  2.71313+. 
Ans,  2.88203+. 

Ans,  0.64549+. 
Ans,  3.20936+. 


135.  Finally,  instead  of  the  last  method,  we  may,  if  we 
please, 

Change  the  common  fraction  into  a  decimal,  and  continue 
the  division  until  the  number  of  decimal  places  is  double  the 
number  of  places  required  in  the  root.  Then  extract  the  root 
of  the  decimal  hy  the  last  rule. 

Exercises. 

1.    Extract  the  square  root  of  —  to  within  .001. 
^  14 

This  number,  reduced  to  decimals,  is  0.785714  to  within  0.000001 ; 
but  the  root  of  0.785714  to  the  nearest  unit  is  .886.     Hence  0.886  is  the 

root  of  —  to  within  .001. 
14 


2.    Find  the  V2|4   to  within  0.0001. 


3.  What  is  the  square  root  of  —  ? 

4.  What  is  the  square  root  of  -  ? 

5 

5.  What  is  the  square  root  of  -  ? 


Ans.  1.6931+. 
Ans.  0.24253+. 

Ans.  0.93541+. 

Ans.  1.29099+. 


ROOTS.  173 

136.  In  order  to  discover  the  process  for  extracting  the 
square  root  of  a  monomial,  we  must  see  how  its  square  is 
formed. 

By  the  rule  for  the  multiplication  of  monomials  (§  42),  we 
have 

{ba'b'cy  -  ba'¥c  X  ba'h'c  =  25a*b'c' ; 

that  is,  in  order  to  square  a  monomial,  it  is  necessary  to 
square  its  coefficient^  and  double  the  exponent  of  each  of  the 
letters.  Hence,  to  find  the  square  root  of  a  monomial,  we 
have  the  following  rule  :  — 

Extract  the  square  root  of  the  coefficient,  for  a  new  coefficient. 
Divide  the  exponent  of  each  letter  by  2,  and  then  annex  all 
the  letters  with  their  new  exponents. 

Since  like  signs  in  two  factors  give  a  plus  sign  in  the  prod- 
uct, the  square  of  —  a,  as  well  as  that  of  +  a,  will  be  +a^ : 
hence  the  square  root  of  a^  is  either  +  a  or  —  a ;  also  the 
square  root  of  25  a^b*"  is  either  +  bab"^  or  —  5a^^  Whence  we 
conclude  that  if  a  monomial  is  positive,  its  square  root  may  be 
affected  either  with  the  sign  +  or  — .  Thus,  V9a*  =  ±  Sa'^ ; 
for  +3a*'^  or  —  Sa'^,  squared,  gives  +9  a*.  The  double  sign 
db,  with  which  the  root  is  affected,  is  read  "  plus  and  minus." 

Exercises. 

1.  What  is  the  square  root  of  64  a^^*? 

\/64a«6*  =  -\-^aW,  for  +  80^62  x  +  Sa^J^  =  +  64a6M  ; 
and  \/64a«F  =  -  8 aW,  for  -  8 a^jz  ><  _  8 a^""  =  +  64 a%^. 

Hence  y/Wcfi¥  =  ±  8  a^i*. 

2.  Find  the  square  root  of  625a^^V.  Ans.  =b  25a5V. 

3.  Find  the  square  root  of  576a*Z>V.  Ans.  ±  24a'^>V. 

4.  Find  the  square  root  of  196 0:^2;*.  Ans.  zb  l^a^yz^. 


174  ELEMENTARY    ALGEBRA. 

5.  Find  the  square  root  of  Ula'b'c''d'\     Arts.  ± 21  a'b'e'd\ 

6.  Find  the  square  root  of  784a^'^>W.     Ans.  zh  28  a'b'c^d. 

7.  Find  the  square  root  of  Sla^b^c^.  Ans.  ±  9a*^V\ 

137.  From  the  preceding  rule  it  follows  that  when  a  mono- 
mial is  a  perfect  square,  its  numerical  coefficient  is  a  perfect 
square,  and  all  its  exponents  even  numbers.  Thus,  25  a^^'"^  is 
a  perfect  square. 

If  the  proposed  monomial  were  negative,  it  would  be  im- 
possible to  extract  its  square  root,  since  it  has  just  been 
shown  (§  136)  that  the  square  of  every  quantity,  whether 
positive  or  negative,  is  essentially  positive.     Therefore 

V^     V^^4^,     V-8a'^5, 

are  algebraic  symbols  which  indicate  operations  that  cannot 
be  performed.  They  are  called  imaginary  quantities,  or  rather 
imaginary  expressions,  and  are  frequently  met  with  in  the 
resolution  of  equations  of  the  second  degree. 

138.  When  the  coefficient  is  not  a  perfect  square,  or  when 
the  exponent  of  any  letter  is  uneven,  the  monomial  is  an 
imperfect  square.  Thus,  98  a5*  is  an  imperfect  square.  Its 
root  is  then  indicated  by  means  of  the  radical  sign.  Thus, 
V98  ab*'.     Such  quantities  are  called  surds  of  the  second  degree. 

Cube  Boot. 

138  a,  The  cube  root  of  a  number  is  one  of  three  equal  fac- 
tors of  the  number. 

To  extract  the  cube  root  of  a  number  is  to  find  a  factor' 
which,  multiplied  into  itself  twice,  will  produce  the  given 
number. 

Thus,  2  is  the  cube  root  of  8,  for  2  X  2  X  2  ==  8 ;  and  3  is 
the  cube  root  of  27,  for  3x3x3  =  27. 


ROOTS. 


175 


Roots    ...  1,     2,      3,       4,        5,         6,         7,        8,        9. 
Cubes    ...  1,     8,     27,     64,     125,     216,     343,     512,     729. 

The  numbers  in  the  first  line  are  the  cube  roots  of  the  cor- 
responding numbers  of  the  second.  The  numbers  of  the  second 
line  are  called  perfect  cubes.  By  examining  the  numbers  of 
the  two  lines,  we  see,  first,  that  the  cube  of  units  cannot  give 
a  higher  order  than  hundreds ;  second,  that  since  the  cube 
of  one  ten  (10)  is  1000,  and  the  cube  of  9  tens  (90)  is  81,000, 
the  cube  of  tens  will  not  give  a  lower  denomination  than  thou- 
sands, nor  a  higher  denomination  than  hundreds  of  thousands. 

Hence,  if  a  number  contains  more  than  three  figures,  its 
cube  root  will  contain  more  than  one  ;  if  it  contains  more  than 
six,  its  root  will  contain  more  than  two ;  and  so  on,  every 
additional  three  figures  giving  one  additional  figure  in  the 
root ;  and  the  figures  which  remain  at  the  left  hand,  although 
less  than  three,  will  also  give  a  figure  in  the  root.  This  law 
explains  the  reason  for  pointing  oflf  into  periods  of  three  fig- 
ures each. 

Let  us  now  see  how  the  cube  of  any  number,  as  16,  is 
formed.  Sixteen  is  composed  of  1  ten  and  6  units,  and  may 
be  written  10  +  6.  To  find  the  cube  of  16,  or  of  10  +  6,  we 
must  multiply  the  number  by  itself  twice. 


To  do  this  we  place  the  number  thus :  - 


Product  by  the  units 
Product  by  the  tens 

Square  of  16 
Multiply  again  by  16 

Product  by  the  units 
Product  by  the  tens 

Cube  of  16 


16  =  10+       6 
10+     6 


100  + 


60+    36 
60 


100+    120+    36 
10+      6 


600+    720  +  216 
1000  +  1200+    360 

1000  +  1800  +  1080  +  216 


176  ELEMENTARY    ALGEBRA. 

By  examining  the  parts  of  this  number,  it  is  seen  that  the 
first  part,  1000,  is  the  cube  of  the  tens;  that  is, 

10  X  10  X  10  -  1000. 

The  second  part,  1800,  is  three  times  the  square  of  the  tens 
multiplied  by  the  units ;  that  is, 

3  X  (10)^  X  6  -  3  X  100  X  6  -  1800. 

The  third  part,  1080,  is  three  tiines  the  square  of  the  units 
multiplied  by  the  tens ;  that  is, 

3  X  6'^  X  10  =  3  X  36  X  10  -  1080. 

The  fourth  part  is  the  cube  of  the  units ;  that  is, 

6^  ==6x6x6  =  216. 

Take,  for  example,  the  following  :  — 
What  is  the  cube  root  of  the  number  4096  ? 

Since  the  number  contains  more  than  .         '    -,rt 

three  figures,  we  know  that   the   root  'lUyD^iO 

will  contain  at  least  units  and  tens.  1 

Separating  the  three  right-hand  fig-    i2wO__q\oa      /q   q   7   a 

ures  from  the  4,  we  know  that  the  cube  V   >     »     > 

of  the  tens  will  be  found  in  the  4,  and  16^  ^=  4  096 

1  is  the  greatest  cube  in  4. 

Hence  we  place  the  root  1  on  the  right,  and  this  is  the  tens  of  the 
required  root.  We  then  cube  1,  and  subtract  the  result  from  4;  and  to 
the  remainder  we  bring  down  the  first  figure,  0,  of  the  next  period. 

We  have  seen  that  the  second  part  of  the  cube  of  16,  viz.,  1800,  is 
three  times  the  square  of  the  tens  multiplied  by  the  units;  and  hence  it 
can  have  no  significant  figure  of  a  less  denomination  than  hundreds.  It 
must  therefore  make  up  a  part  of  the  30  hundreds  above.  But  this  30 
hundreds  also  contains  all  the  hundreds  which  come  from  the  third  and 
fourth  parts  of  the  cube  of  16.    If  it  were  not  so,  the  30  hundreds,  divided 


ROOTS.  177 

by  three   times   the  square  of  the   tens,   would  give  the   unit   figure 
exactly. 

Forming  a  divisor  of  three  times  the  square  of  the  tens,  we  find  the 
quotient  to  be  10 ;  but  this  we  know  to  be  too  large.  Placing  9  in  the 
root  and  cubing  19,  we  find  the  result  to  be  6859.  Then  trying  8,  we 
find  the  cube  of  18  still  too  large;  but  when  we  take  G,  we  find  the 
exact  number.     Hence  the  cube  root  of  4096  is  16. 

Hence,  to  find  the  cube  root  of  a  number, 

Separate  the  given  number  into  periods  of  three  figures  each^ 
by  placing  a  dot  over  the  place  of  units,  a  second  over  the  place 
of  thousands,  and  so  on  over  each  third  figure  to  the  left.  The 
left-hand  period  will  often  contain  less  than  three  places  of 
figures. 

Note  the  greatest  perfect  cube  in  the  first  period,  and  set  its 
root  on  the  right  after  the  manner  of  a  quotient  in  division. 
Subtract  the  cube^ of  this  number  from  the  first  period,  and  to 
the  remainder  bring  down  the  first  figure  of  the  next  period  for 
a,  dividend. 

Take  three  tijnes  the  square  of  the  root  just  found  for  a  trial 
divisor,  and  see  how  often  it  is  contained  in  the  dividend,  and 
place  the  quotient  for  a  second  figure  of  the  root.  Then  cube 
the  figures  of  the  root  thus  found,  and,  if  their  cube  be  greater 
than  the  first  two  periods  of  the  given  number,  diminish 
the  last  figure;  but  if  it  be  less,  subtract  it  from  the  first  two 
periods,  and  to  the  remainder  bring  down  the  first  figure  of  the 
next  period  for  a  new  dividend. 

Take  three  times  the  square  of  the  whole  root  for  a  secoiid 
trial  divisor,  and  find  a  third  figure  of  the  root.  Cube  the 
ivhole  root  thus  found,  and  subtract  the  result  from  the  first 
three  periods  of  the  given  number  when  it  is  less  than  that 
number,  but  if  it  is  greater,  diminish  the  figure  of  the  root. 
Proceed  in  a  similar  way  for  all  the  pei'iods. 

D.  >-.  E.  A.  —  12. 


li. 


178  ELEMENTARY    ALGEBKA. 

Take  the  following  example  :  — 

What  is  the  cube  root  of  99252847  ? 

99  252  847 1 463 
4^=         64 


4^X3=    48 1  352  1st  dividend. 

First  two  periods  99  252 

(46)^  ==  46  X  46  X  46  =         97  336 

■  3x(46)'^  =  6348|l9168     2d  dividend. 

First  three  periods  99  252  847 

(463)^  -         99  252  847 

Exercises. 

Find  the  cube  roots  of  the  following  numbers  :  — 

1.  Of  389017.  '  4.   Of  84604519. 

2.  Of  5735339.  5.    Of  259694072. 

3.  Of  32461759.  6.    Of  48228544. 

138  b.    To  extract  the  cube  root  of  a  decimal  fraction, 

An7iex  ciphers  to  the  decimal,  if  necessary,  so  that  it  shall 
consist  of  three,  six,  nine,  etc.,  places.  Then  put  the  first  point 
over  the  place  of  thousandths,  the  second  over  the  place  of  mil- 
lionths,  and  so  on  over  every  third  place  to  the  right ;  after 
which  extract  the  root  as  in  whole  numbers. 

Notes.  —  1.  There  will  be  as  raany  decimal  places  in  the  root  as 
there  are  periods  in  the  given  number. 

2.  The  same  rule  applies  when  the  given  number  is  composed  of  a 
whole  number  and  a  decimal. 

3.  If  in  extracting  the  root  of  a  number  there  is  a  remainder  after  all 
the  periods  have  been  brought  down,  periods  of  ciphers  may  be  annexed 
by  considering  them  as  decimals. 


ROOTS.  179 

Exercises. 
.  Find  the  cube  roots  of  the  following  numbers  :  — 

1.  Of  .167464.  4.   Of  .751089429. 

2.  Of  .870983875.  5.   Of  .353393243. 

3.  Of  12.977875.  6.   Of  3.408862625. 

138  c.    To  extract  the  cube  root  of  a  common  fraction, 

Reduce  compound  fractions  to  simple  ones,  mixed  numbers 
to  improper  fractions,  and  then  reduce  the  fraction  to  its  lowest 
terms. 

Extract  the  cube  root  of  numerator  and  denominator  sepa- 
rately if  they  have  exact  roots ;  but,  if  either  of  them  has  not 
an  exact  root,  reduce  the  fraction  to  a  decimal,  and  extract  the 
root  as  in  the  last  case. 

Exercises. 
Find  the  cube  roots  of  the  following  fractions  :  — 

1.  Of  Iff.  3.    Of^V  5.    Off 

2.  Of  31^1^.  4.    Off  6.    Off 

Any  Boot. 

138  d.  To  find  a  rule  for  extracting  any  root  of  a  monomial, 
we  have  simply  to  reverse  the  rule  of  §  119  for  finding  any 
power  of  a  monomial.  This  gives  for  the  extraction  of  any 
root  of  a  monomial  the  following  rule  :  — 

Extract  the  required  root  of  the  coefficient,  for  <t  new  coeffi- 
cient. 

Divide  the  exponent  of  each  letter  by  the  index  of  the  root. 


180  ELEMENTARY    ALGEBRA. 

Exercises. 
Extract  the  indicated  roots  of  the  following  monomials :  — 


1.  </64:a'b'c\  Ans.  +^a'bc\ 

2.  -v/lGa^^V.  Ans.  ±2a'b'c. 

Note.  —  It  may  be  shown,  as  in  §  119,  that  any  power  of  an  even 
degree  of  a  quantity  is  +  :  hence  every  root  of  an  even  degree  must 
have  the  double  sign  ±.  Every  power  of  an  odd  degree  of  a  quantity 
has  the  same  sign  as  the  quantity :  hence  every  root  of  an  odd  degree 
of  a  quantity  must  have  the  same  sign  as  the  quantity. 


3.    V-8a^ 


4.  V-32a^W. 

5.  </Sl  a'b'\ 


6.  V8a«6Vl 

7.  A/81a*6V«. 


8.    V- 32 aV«c^^^ 


A^ 

as.  —2  a. 

Ans 

1.  -2a'b. 

Ans 

r.  ±Sab\ 

Ans. 

+  2a'bc\ 

Ans. 

±3ab'c'. 

Ans. 

-2ac'd\ 

Ans. 

-ba'b'c. 

9.    V~125a^Z>V. 

Ani/  root  of  an  even  degree  of  a  negative  quantity  is 
imaginary ;  for  there  is  no  quantity  which,  raised  to  a  power 
of  an  even  degree,  will  give  a  negative  result.     Thus, 

are  imaginary  quantities. 

When  the  coefficient  is  not  a  perfect  power  of  the  degree 
indicated,  or  when  the  exponent  of  any  letter  is  not  exactly 
divisible  by  the  index  of  the  root,  the  monomial  is  an  imper- 
fect power  of  the  degree  indicated.  Thus,  IQa'^b*  is  an  im- 
perfect cube,  and  210;?/^  is  an  imperfect  fourth  power.  Their 
roots  are  indicated  thus  :  — 


■Vl6a'b\     </2lx7/'. 


ROOTS.  181 

Such  quantities  are  called  surds.  Hence  a  surd  is  the 
indicated  root  of  an  imperfect  power  of  the  degree  indicated. 

138  e.  For  the  rule  to  extract  any  root  of  a  fraction,  reverse 
the  rule  given  in  §  120  for  finding  any  power  of  a  fraction, 
and  we  have  the  rule  :  — 

Extract  the  required  root  of  the  numerator^  for  a  neiu 
numerator ;  and  the  required  root  of  the  denominator,  for  a 
new  denominator. 

Exercises. 
Find  the  indicated  roots  of  the  following  fractions :  — 


i 


/^  .         2ay 

—  Ans.  ^• 

21  dh'  3c^V 


2.    V|l^'.  ^n5.  ±^. 

\6252'  bz' 


3 1 343  a;-^-^ 


\16a;y* 

si     Ifc^x^"^ 
\       32y^V«* 

'    M     1285V* 


. 

7x-'y-' 

a-'b-'c' 

Ans. 

ah'c 
-  2x'f 

Ans. 

Wx"^ 

2y^z^ 

Ans 

axif 
•       2hz'' 

182  elementary  algebra. 

Radicals. 

Transformation  of  Radicals. 

139.  Let  a  and  b  denote  any  two  numbers,  and  p  the  prod- 
uct of  their  square  roots.     Then 

^aX-\/b=p  (1) 

Squaring  both  members,  we  have 

axh  -=f  (2) 

Extracting  the  square  root  of  both  members  of  (2),  we  have 

V^-J9  (3) 

Since  the  second  members  are  the  same  in  Equations  (1) 
and  (3),  the  first  members  are  equal ;  that  is. 

The  square  root  of  the  product  of  two  quantities  is  equal  to 
the  product  of  their  square  roots. 

Similarly,  if  n  denotes  any  number,  then  n Va  will  denote  the 
nth  root ;  that  is,  any  root  of  a.  Let  I  denote  the  products 
of  the  nth  roots  of  a  and  b.     Then 

^axVb^l  (1) 

Raising  both  members  to  the  nth  power,  we  have 

axb^l""  (2) 

Extracting  the  nth  root  of  both  members  of  (2),  we  have 

^^=1  (3) 

Since  the  second  members  are  the  same  in  Equations  (1) 
and  (3),  the  first  members  are  equal ;  that  is, 

The  nth  root  of  the  product  of  two  quantities  is  equal  to  the 
product  of  their  nth  roots. 


RADICALS.  183 

140.   Let  a  and  b  denote  any  two  numbers,  and  q  the  quo- 
tient of  their  square  roots.     Then 

Squaring  both  members,  we  have 

f=?'  (2) 

Extracting  the  square  root  of  both  members  of  (2),  w^e  have 

Since  the  second  members  are  the  same  in  Equations  (1) 
and  (3),  the  first  members  are  equal ;  that  is, 

The  square  root  of  the  quotient  of  two  quantities  is  equal  to 
the  quotient  of  their  square  roots. 

Similarly,  if  r  denotes  the  quotient  of  the  nth  roots  of  a  and  , 
h,  we  have 

^  =  r  (1) 

Raising  both  members  to  the  nth  power,  we  have 

f  =  ^"  (2) 

Extracting  the  nth  root  of  both  members  of  (2),  we  have 


4t'         .      <') 


Since  the  second  members  are  the  same  in  Equations  (1) 
and  (3),  the  first  members  are  equal ;  that  is, 

The  nth  root  of  the  quotient  of  two  quantities  is  equal  to  the 
quotient  of  their  nth  roots. 


184  ELEMENTARY    ALGEBRA. 

These  principles  enable  us  to  transform  radical  expressions, 
or  to  reduce  them  to  simpler  forms.     Thus,  the  expression 

98a5*  =  49Z>*x2a. 
Hence  V98aF  =  V49Fx~2a  ; 

and,  by  the  principle  of  §  139, 

V49^>*  X2a  =  V49F  X  V2^=  1h^^/2a, 
In  like  manner, 


V45  a'b^c'd  =  ^dd'b'c'x^bd      =  3  abcV^bd. 


V864a'^Z)V^  =  -VlUd'b'c''  xUc=  12  ab'cW 6  be. 
Similarly, 

V54:xyz      =  •^27^yx2y2       =  3  xf  V2^z. 

The  coefficient  of  a  radical  is  the  quantity  without  the  sign. 
Thus,  in  the  expressions 

Ib'-^^^',     SabcVbbd,     12  ab'c'-\/6bi, 

the  quantities  *Jb^,  Sabc,  12aZ)V,  are  coefficients  of  the  radicals. 

141.  Hence,  to  simplify  a  radical  of  the  second  degree,  we 
have  the  following  rule  :  — 

jResolve  the  expression  under  the  radical  sign  into  two 
factors,  one  of  which  shall  be  a  perfect  square. 

Extract  the  square  root  of  the  perfect  square,  and  then  multi- 
ply this  root  by  the  indicated  square  root  of  the  remaining 
factor. 

In  like  manner,  to  simplify  a  radical  of  the  nth  degree, 
that  is,  of  any  degree,  we  have  the  rule  :  — 

Resolve  the  expression  under  the  radical  sign  into  two  factors, 
one  of  which  shall  be  a  perfect  nth  power. 

Extract  the  nth  root  of  the  perfect  nth  power,  and  then 
TJiultiply  this  root  by  the  indicated  nth  root  of  the  remaining 
factor. 


'  RADICALS.  185 

Exercises. 
Eeduce  the  following  radicals  to  their  simplest  form  :  — 


1. 

■\/lba'bc. 

2. 

■V12S  b'a'd\ 

3. 

VS2a'b'c. 

4. 

■V256a'd'c\ 

5. 

■Vl024:aWc\ 

6. 

■V129  a'b'c'd. 

7. 

V67ba'b'c'd. 

8. 

Vl445aV^*. 

9. 

VIOOS  a'dhn\ 

10. 

■V2lb6a'%'c\ 

11. 

■V^ODa'b'd\ 

12. 

■</48a^5V. 

13. 

■^  112  a^b\ 

14. 

2-v/54a;-V'. 

15. 

■</bl2a'y-\ 

16. 

Si^l2SxY'' 

Atis.  ba-y/Sabc, 

Ans.  Sb'a'dV2b, 

Ans,  4:a*b*\^2ac, 

Ans.  16  ab^c*. 

A71S.  32a'b'c'Vabc. 

Ans.  27a'b'cWM. 

Ans.  15a'b^cVSabd. 

Ans.  17acW^V6a. 


Ans.  12  a'd'mWTad. 

Ans.  Ua'b'c'Vn. 

Ans.  9aWd*Vba, 

Ans.  2acV3a2^. 

Ans.  2a5Wl4a. 


Ans.  6x-hj-W2x-\ 
Ans.  Aa"'7j-'^^/2y~\ 


Ans.  12  xy'^-yj  2  x^y~'^. 
17.    -\/-24:i'-y.  Ans.  --2x-^y^Zf. 

142.  A  coefficient,  or  a  factor  of  a  coefficient,  may  be  carried 
under  the  radical  sign,  raising  it  to  the  proper  power,  by 
squaring  it  if  the  radical  is  of  the  second-  degree,  by  cubing 
it  if  the  radical  is  of  the  third  degree,  by  raising  it  to  the 


18(3 


ELEMENTARY    ALGEBRA. 


fourth  power  if  the  radical  is  of  the  fourth  degree,  and  so  on. 
Thus, 


(1)  ^a'Vhc  -  V(3a'7  xbc^  y/^a'hc. 

(2)  2ahVd  =  2^~^^^¥d^'\/^^c^m. 

(3)  4(a  +  5)Va'^=T  =  4V(a+^)Xa-^>)==4V(a^-60(a  +  6}. 


(4)  5  hcy/a^  -c^  =  bVb'c'  {a'  -  c'). 


(5)  3a^/2ax  =  a/(3  af  x2ax=  V27  aJ'x2ax=  V54  a'x. 

(6)  -  4  v"^  =  </i-4:yxab'  =  </2M^\ 


(7)  3 ab-y/¥^'  =  S-Va'b' (x'  -  y'). 

(8)  2x'</^r^'  =  -y/lQx^a'-b'). 

The  square  root  of  a  negative  quantity  may  be  simplified. 


Thus, 


/4a'  X 


V^  =  V9X-1  =  V9  X  V^^  =  3-> 

2^  =  2aV^^^ 
=  2aV2?  X  V^ ;  that  is. 


■1, 


and       V—  4  d^ 

also       V— 8a'^  =  V4a' X 


^Ae  square  root  of  a  negative  quantity  is  equal  to  the  square 
root  of  the  same  quantity  with  a  positive  sign,  multiplied  into 
the  square  root  of  —1. 


Exercises. 


Eeduce  the  following  :  • 
1.    V-64a*^^)l 


2.    V-128a*6*. 


3.  V-72a^^>V. 

4.  V^^^^^48a^. 


Ans.  SabV—  1. 

Ans.  Sa'bW2b'V^^. 

Ans.  6a'b'cW2^V^^, 

Ans.  iac^VSabcV—  1. 


RADICALS.  '  187 

Note.  —  Any  even  root  of  a  negative  quantity  is  equal  to  the  same 
root  of  the  quantity  with  a  positive  sign,  multiplied  into  the  same  indi- 
cated root  of  —  1. 

5.  ■</'-  Slxy  =  -y/STxy  X  V"^  =  3xy\/^^. 

6.  V'-  32 a'b'c  =  </S2  a'b'  X  ab'c  X  V=l  =  2ab y/'^'V^, 

142  a.  It  is  often  desirable  to  change  the  index  of  a  radical, 
and  it  may  be  done  in  accordance  with  the  following  prin- 
ciple :  — 

Let  m  and  n  represent  any  two  numbers  whatever,  and  let 

^i'^  =  s  (1) 

Kaise  both  members  of  (1)  to  the  w.th  power,  and  we  have 

■Va  -  s"*  (2) 

Raise  both  members  of  (2)  to  the  nth  power,  and  we  have 

a  =  s"*^  (3) 

Extract  the  mnth  root  of  the  members  of  (3),  and  we  have 

"Va-5  (4) 

Since  the  second  members  of  Equations  (1)  and  (4)  are  the 
same,  the  second  members  are  equal,  and  we  have 

that  is. 

The  mth  root  of  the  nth  root  of  a  quantity  is  equal  to  the 
mnth  root  of  that  quantity,  and  the  reverse, 

and  i^  ^  yj'^^ 

Take  -^io^. 

By  this  principle  we  have 

-VI^'  -  -y  V4^  ; 

OF  THE  ^ 

i     UNIVERSITY  ] 


188  ELEMENTARY    ALGEBRA. 

and  since  V4a"''  =  2  a, 

we  have  ■\/'4:d^  =  a/2  a. 


In  like  manner      VMa^'  -  ^^-y/SQd'b'  =  -VQab, 
and  generally  "VF  -  >/ W  =  VJ  ; 

that  is,  when  the  index  of  a  radical  can  be  divided  by  n,  and 
the  quantity  under  the  radical  sign  is  a  perfect  nth  power, 
we  can  divide  its  index  by  n  provided  we  extract  the  nth  root 
of  the  quantity  under  the  sign. 

Thus,  •</27m'Vy  =  ^'ErrMp  ; 


-VS^^=-y/2xy^; 


The  converse  of  this  principle  is  true.     For  example : 
2a=  V4a^ 


Hence  -^2^  =  ^|  VIS^  =  -y/I^'. 

Here  the  radical  ■\/2a  has  been  changed,  without  altering 
its  value,  by  multiplying  the  index  by  2  and  at  the  same  time 
raising  the  quantity  under  the  radical  sign  (2a)  to  the  second 
power.  As  all  other  examples  can  be  treated  in  the  same 
way,  we  have  the  principle  :  — 

The  index  of  a  radical  may  be  multiplied  by  any  number^ 
provided  we  raise  the  quantity  under  the  radical  sign  to  a 
power  of  which  this  number  is  the  exponent. 

Thus,  -v/4^=-</T6^; 

V2  =  -v/l6; 


si        1  9|        1  18[ 


64* 


RADICALS.  189 

The  same  principles  are  illustrated  when  the  radical  is 
denoted  by  a  fractional  exponent.     Thus, 

(a)* -(a)*,     since  1  =  I ; 
also  (ay'^  =  {a)\  since  y^  =  \] 

that  is,  a*  —  a^\     and    a^'^  =  a^  -^ 

or,  by  writing  the  equivalent  expressions, 

Va  =  ■^/a^         Va^  =^-  V«, 
the  same  results  as  obtained  before  under  the  rules.     Hence, 
in  the  case  of  fractional  exponents,  we  may 

Multiply  or  divide  both  terms  of  the  fractional  exponent  by 
the  same  number  without  changing  the  value  of  the  radical. 

By  means  of  these  principles,  radicals  having  different 
indices  may  be  reduced  to  equivalent  radicals  having  a 
common  index.  Take  the  radicals  Va,  V(?,  and  VJ.  The 
L.  0.  M.  of  the  indices  2,  3,  and  4,  is  12.  Reducing  each  to  the 
index  12  by  the  foregoing  principle,  we  have 

Va  —  Va^,    Vc  =  Vc?*,    and    V^  =  V^^. 

As  all  other  cases  may  be  treated  in  the  same  way,  we  have 
the  rule  :  — 

Find  the  L.  C.  M.  of  the  indices,  and  reduce  each  radical  to 
that  index. 

When  the  radicals  are  expressed  by  fractional  exponents, 
we   have   simply   to   reduce   these    exponents  to  a  common 

denominator.     Thus,  a^,  5^  c^ ,  and  d^  become,  when  reduced 

to  a  common  index   6,  a^  b'^\  c^,  and  d^  ;  or  (a^)*,  (^^'0  > 

(cO*,  and  {d'f. 

In  like  manner  reduce  2,  (3)^,  {ay,  and  {by  to  a  common  index. 

Ans,  (2")T^,  (3*)T^  (a«)TV,  (^,3)tV  .  or  V2,  V3\  V^,  Vb\ 


190  ELEMENTARY    ALGEBRA. 


Addition  of  Hadicals. 

143.  Similar  radicals  are  those  which  have  a  common  unit 
or  radical  part.  Thus,  the  radicals  SVJand  5cV^  are  simi- 
lar, and  so  also  are  9 V2  and  7 V2  ;  ^Vc,  aVc,  and  —  3v  c 
are  similar,  as  are  VSa,  9V3a,  TcVSa,  and  xyA/Za. 

144.  Kadicals  are  added  like  other  algebraic  quantities. 
Hence  the  following  rule  :  — 

If  the  radicals  are  similar^  add  their  coefficients^  and  to  the 
sum  annex  the  common  unit  or  radical. 

If  the  radicals  are  not  similar^  connect  them  together  with 
their  proper  signs. 

Thus,       3  a^l  +  5  c^l  --  (3  a  +  5  c)Wb. 
In  like  manner 

7 V2^+  3 V2^-  (7  +  3)V2^---  10Vi2^. 

a^2l  +  hV2l  -  (a  +  h)-s/21. 

Notes.  —  1.  Two  radicals,  which  do  not  appear  to  be  similar  at  first 
sight,  may  become  so  by  transformation  (|J  141,  142  a).     Thus, 


V48  ah"  +  6V75a  =  45\/3^  +  56\/3^  =  96\/3a. 

2\/45  +  3  V5  =  6\/5 +•  3  V5  =  9\/5. 

3v/4^2  _^  2\/2^=  3v/2^  +  2v/2^  =  5\/2^. 

5\/8  +  4  v^5l2  =  5  V8  +  4\/8  =  9  Vs. 

2.  When  the  radicals  are  not  similar,  the  addition  or  subtraction  can 
only  be  indicated.  Thus,  in  order  to  add  3V6  to  by/ a,  we  write 
5Va  +  3V6;  and  to  add  2\/c  and  3Vc,  we  write 

2  v/c  +  3  v^c,  or  2  v^?  +  3  v^?. 


RADICALS.  191 


Exercises. 
Add  the  following  :  — 


1.  V27a'  and  V48a^  Ans.  7aV3. 

2.  V50a^  and  V72a^.  Ans.  lla'bV^ 

3.  ^^-f  and  ^^.  ^ns.  4a^l. 


4.    Vl25  and  VSOOa'^  Ans.  (5+10a)V5. 

I'W     .    /Too  .      10  /^ 


6.  V98^  and  V36l?  -  36  a^    ^ns.  7aV2i  + 6Va;-^-al 

7.  V98a^  and  V288aV.  ^725.  (7a  +  12aV)V2a;. 

8.  V72  and  Vl28.  Ans.  14V2. 

9.  V27  and  VI47.  Ans.  10V3. 

10.  \/-  and  \~  Ans.   —  V6. 
^3            >'50  •                                 30 

11.  2VaFb  and  3V646^.  Ans.  (2 a  +  24: x') Vb . 

12.  V243  and  10V363.  Ans.  119V3. 


13.  V320^  and  V245VF.  ^ris.  (8aZ>  +  7a'Z/')V5. 

14.  V75^^  and  V300^^.         Ans.  (5  a'b' +10  a'b')\'U. 

15.  3  5a/2^^,  7^2^^and8a^2^.    Ans.  IS abi^J^'. 

16.  -v^8 a^^)  +  16 a'  and  a/Z>*  +  2 a^>^    ^ns.  (2a  +  ^)-v/2^+T. 

17.  6^4^,  2-</2^,  and  ^^8^.  ^715.  9^/2^. 


192  ELEMENTARY   ALGEBRA. 


18.    12v'i  and  S-yi-- 

>'4       ^8  8      2 

Hence  12a/1  =  l?v^  =  6v/2, 

^4       2 

^32       >64       >        64      4 

Hence  3-\/— =  ?v/2. 

^32     4 

3\9     6\36  5\32  5^^* 

Subtraction  of  Radicals. 

145.  Radicals  are  subtracted  like  other  algebraic  quantities. 
Hence  the  following  rule  :  — 

If  the  radicals  are  similar,  subtract  the  coefficient  of  the 
subtrahend  from  that  of  the  minuend,  and  to  the  difference 
annex  the  common  unit  or  radical. 

If  the  radicals  are  not  similar,  indicate  the  operation  by  the 
mi7ius  sign. 

Exercises. 

1.    "What  is  the  difference  between  3aV5  and  a^/bf 

Here  3ay/b  —  ay/h  =  2ay/h.     Aiis. 


2.    From  ^a^21b'  subtract  6aV27F.  Ans.  9a5V3. 

9a\/27i2  =  27 a5\/3,  and  GaV27F=  18a^/V3. 
2ra6\/3  -  18a6\/3  =  9a6>/3. 


RADICALS.  193 

Find  the  differences  between  the  following  :  — 
3.    V75  and  V48.  Arts.  VS. 


4.    -\/24:a'b'  and  V545\  Ans.  (2ab-Sb')V6. 

6.    VT28^^  and  V32a',  Ans.  (8a5  -  4a*)V2^. 


7.  -V4:8a'b'  and  Vdab,  Ans.  AabVSab  -  SVab. 

8.  ■V2i2¥l?  and  V2a'b\  Ans.  (lla'b' -ab)V2^, 

10.    V32aa^  and  VSOc?,  ^ns,  4aV5. 


11.    V720a'^'  and  V24:6abc'd.      Ans.  (I2ab -7cd)V^ab. 


12.    VdQSd'b'  and  V200a^Z>^  ^tis.  12a^V2. 


13.  Vll2aV  and  V28a«6l  ^/is.  2a'bW7. 

Note.  —  Two  radicals  which  do  not  appear  at  first  to  be  similar  may 
become  so  by  transformation,  as  in  the  preceding  case.     For  example : 

\/216a66*  -  by/WM  =  \/2mM^  xb-  hy/21  a^  x  h 
=^Q>a}h\/b  -Za'bx/b  =  Sa^by/b, 
and  \/320  -  V80  =  8 \/5  -  4>/5  =  4>/5. 

14.  From  bV^'  subtract  3\/2^,  Ans.  2^2a. 

15.  From  -v/8l  +  -v/l92  subtract  a/5T2.         Ans.  7\/3-8. 

Note.  —  Here  V512  is  not  similar  to  the  sum  of  >/8l  and  Vl92; 
hence  the  operation  of  subtraction  can  only  be  indicated. 


,^      -c,  2    32,333        r ,       ,  1    3   1      .  \         31   8/^ 

16.    From  - X- +  - a^t;  subtract  - -vh^'  -^^^•x;;v6. 

3  \9     5  \32  6  \36  90 


•  13. 


194  ELEMENTARY    ALGEBRA. 

Multiplication  of  Radicals. 

146.    Radicals  are  multiplied  like  other  algebraic  quantities. 
Hence  we  have  the  following  rule  :  — 

Multiply  the  coefficients  together,  for  a  new  coefficient. 
Multiply  the  radical  parts  together,  for  a  new  radical  part. 
Then  reduce  the  result  to  its  simplest  form. 

Exercises, 

1.  Multiply  3aV5?  by  2Va?. 

3  aVhc  X  2y/ah  =  3  a  x  2  x  Vbc  x  Va6, 
which,  by  §  139,  =  Qas/Wac  =  6 ah\/ac. 

Multiply  the  following  :  — 

2.  3V5a6  and  4V20a.  .4ns.   120aV^. 

3.  2a-y/bc  and  SaV^c?.  Ans.  Qa^bc. 


4.    2 a-yjd'  +  &  and  -  3  a Va'  +  h\        Ans.  —  6 a\a^  +  b""). 


5.  2ab-\/a  +  b  and  ac^a  —  b.  Ans.  ^a^bc^d^  —  b'\ 

6.  3V2  and  2V8.  Ans.  24. 

7.  |^|a^^  and  A^|,.^.  ^,,.  ^a^.VlT. 

8.  2a;  +  VJ  and  2a;  -  VJ.  .47is.  4a;'  --  6. 


9.    ^^a  +  2  VJ  and  ^^a  -  2 V^.  ^ns.  Va'  -  4  6. 

10.    3aV27V  and  V2a.  Ans.  9aV6. 

Note.  —  When  the  radicals  have  not  the  same  index,  they  must  be 
reduced  to  a  common  index  before  multiplying.  For  example  :  to  mul- 
tiply 2aVx  by  3\/y,  we  must  reduce  them  to  the  common  index  6, 
making  them  2av^  and  3\^  :  and  the  product  is  Oav^.r'y*. 


RADICALS.  195 

11.  V8  and  \/5.      Ans.  ^5l2  X  ^25  -  -v''r2800  -  2</M). 

12.  3a>/6and55v^2^.  Ans.  15ab\^W?. 

13.  3aA/8a^  and  S^a/JoV.     ^ns.  6a5v^32a*c  =  12a'5\^2c. 

Note.  —  By  combining  the  foregoing  method  with  the  method  for  the 
multiplication  of  polynomials  (J  43),  complicated  radical  expressions  may 
be  multiplied  together. 

14.  \/i  +  2-v/S  +  4  and  Vx-^2</x, 

First  Method.  Second  Method. 


^/^+2^; 


X 


a:*  +  2a;* 

a;^  +  2a;^  +  4a;* 

ri 

'x'  +  2-v/a:^  +  4</a; 

+  2^?  +  4\/i  +  8A/i 

Note,  —  The  two  results  are  identical,  but  the  second  has  been 
obtained  by  following  the  ordinary  rule  for  exponents.  Hence  we 
conclude  that  the  rule  for  multiplication  is  the  same  whether  the  expo- 
nents are  entire  or  fractional. 

15.    (a*  +  aM  +  o^x  +  x^)  X  (a*  -  :^).  Ans.  a  -  x\ 


16.    (:r+|  + 


■V» +!>(.■ +I-V.+0 


Ayis.  x^  -{-px  —  q. 

"■  (lN/fW5)x(fVf-Vl}  ^-f'-£ 

18.    (\'a--^  +  ^^ah)  X  (yjoT^  -  ^lah).         Ans.  "v/^^  -  yjah. 


xe.     l+iV^r-^   X   l+^V^ 


4      2 


196  ELEMENTAEY    ALGEBRA. 

Division  of  Radicals. 

147.  Radical  quantities  are  divided  like  other  algebraic 
quantities.     Hence  we  have  the  following  rule  :  — 

Divide  the  coefficient  of  the  dividend  hy  the  coefficient  of  the 
divisor y  for  a  new  coefficient. 

Divide  the  radical  part  of  the  dividend  hy  the  radical  part 
of  the  divisor^  for  a  new  radical  part. 

Then  reduce  the  result  to  its  simplest  form. 

Exercises. 

1.  Divide  SaVb^c  by  4iay/bc^. 

—  =  2,  New  coefficient. 
4a 

Hence  2  x  -  =  ^^.    Quotient. 

c       c 

2.  Divide  baVb  by  2b^c.  Ans.  ^\' 

3.  Divide  12acV6^  by  4cV2T.  Atis.  SaVSc. 

4.  Divide  6aV966*  by  SVsF.  Ans.  4:abVS. 

5.  Divide  4a^V50P  by  2aV5j.  '  Ans.  2bWlO. 

6.  Divide  2Ga'b^81d'¥  by  13aV9^.  Ans.  6a'by/ab. 

7.  Divide  S4.a'bW27ac  by  42a5V3a.  ^ns.  6a'6'Vc. 

6.    Divide  J^a'  by  V2.  Ans.  ]a. 

'  o  4 

9.    Divide  6a^5V20^  by  12V5a.  .Ins.  a'b\ 


RADICALS. 

•197 

10. 

Divide  6aVl06''  by  3V5. 

^Ms.  2oJV2. 

11. 

Divide  48J*Vl5  by  26'^- 

.4m.  360  Z.'. 

12. 

Divide  Ba'b*cWld'  by  2a-\/2id. 

Am.  2a¥<?d. 

13. 

Divide  96aVV986=^  by  48aicV26. 

Ans.  14  a'6c*. 

14. 

Divide  27a'6V21a'  by  V7a. 

Ans.  27aWV3. 

15. 

Divide  IBaWVSa'  by  &abyjd\ 

^«s.  6a«6=V2. 

Note.  —  As  in  multiplication,  if  the  radicals  have  not  the  same  index, 
they  must  be  reduced  to  a  common  index  before  division. 

16. 

Divide  2V2aa;  by  V46a;l 
,       2V2ax  =  2v'8aV, 

and 

>y46x'^  =  v^ieftV. 

17.    Divide  i  a/2  by  a/3. 
2i 

^'^^^  2\81- 

18.  Divide  2V3a  by  'v^4a6.  ^ns.  ^-Oy^- 

Note.  —  By  combining  the  above  methods  with  the  method  for  the 
division  of  polynomials  (§  52),  any  complicated  radical  expression  may 
be  divided  by  another. 

19.  Divide  x  +  Vxt/  +  y  by  Vx  +  -y/xy  +  Vy. 

First  Method. 


x+^xy  +y 
X  +  V.'T^y  +  ■\/xy 


y/x  +  -y/xy  +  -y/y 


■y/x  —  -V^rry  +  Vy 


—  -y/x^y  —  s/xy  —  -\/xy^ 

Vxy  +  i/xy""  +  y 
-Vxy^  Vxy^  +  y 


198  ELEMENTARY    ALGEBRA. 


Second  Method. 


X  +  x^y^  +  x^y^ 


xi  4-  x^y^  +  y^ 


xi  —  x^y^  +  y* 


—  x^ir  —  ^  V   —  x^i 


a 


0 

Note.  —  The  two  results  are  identical,  but  the  second  one  has  been 
obtained  by  following  the  ordinary  rule  for  exponents  :  hence  we  con- 
clude that  the  operation  for  division  is  the  same,  whether  the  exponents 
are  entire  or  fractional. 


...  (ie.-g).(2,t-|). 


Am,  8:r*  +  2a;*y  +  ^^V  +  ^3/'- 

Square  Root  of  Polynomials. 

148.  Before  explaining  the  rule  for  the  extraction  of  the 
square  root  of  a  polynomial,  let  us  first  examine  the  squares 
of  several  polynomials.     We  have 

(a  +  Z)  +  c)'  =  a'^  +  2a^>  +  ^'  +  2(a  +  ^)  d?  +  c^ 
(a  +  5  +  c  +  c?)^  =  a^  +  2  a5  +  ^'  +  2  (a  +  ^))  c  +  c' 

+  2(a  +  5  +  c)c/+cf. 

The  law  by  which  these  squares  are  formed  can  be  enum- 
erated thus : — 

The  square  of  any  'polynomial  is  equal  to  the  square  of  the 
first  term,  plus  twice  the  product  of  the  first  term  hy  the  second, 


ROOTS   OF    POLYNOMIALS.  199 

plus  the  square  of  the  second ;  plus  twice  the  sum  of  the  first 
two  terms  multiplied  hy  the  third,  plus  the  square  of  the  third ; 
plus  twice  the  sum  of  the  first  three  terms  m^ultiplied  hy  the 
fourth^  plus  the  square  of  the  fourth ;  and  so  on. 

149.  Hence,  to  extract  the  square  root  of  a  polynomial,  we 
have  the  following  rule  :  — 

Arrange  the  polynomial  with  reference  to  one  of  its  letters^ 
and  extract  the  square  root  of  the  first  term.  This  will  give  the 
first  term  of  the  root. 

Divide  the  second  term,  of  the  polynomial  hy  double  the  first 
term  of  the  root^  and  the  quotient  will  he  the  second  term  of  the 
root. 

Then  form  the  square  of  the  algebraic  sum  of  the  two  terTns 
of  the  root  found,  and  subtract  it  from  the  first  polynomial ; 
and  then  divide  the  first  term  of  the  remainder  hy  double  the 
first  term  of  the  root ;  and  the  quotient  will  he  the  third  term,. 

Form  the  double  product  of  the  sum  of  the  first  and  second 
terms  by  the  third,  and  add-  the  square  of  the  third ;  then  sub- 
tract this  result  from  the  last  remainder,  and  divide  the  first 
term  of  the  result  so  obtained  by  double  the  first  term  of  the 
root;  and  the  quotient  will  be  the  fourth  term. 

Then  proceed  in  a  similar  munner  to  find  the  other  terms. 

Exercises. 
1.    Extract  the  square  root  of  the  polynomial 

49  a^h^  -  24  ah^  +  25  a*  -  30  a^b  +  l^  b\ 
First  arrange  it  with  reference  to  the  letter  a. 


25  a*- 

-30a36+    da'b-' 

-24a6«  +  166* 

ba'-Sab  +  ib^ 

25  a*- 

10  a' 

40a*^62_ 
AOa^b^- 

-24a6^  +  16  6* 
0 

1st  rem. 
2d  rem. 

200  ELEMENTARY    ALGEBRA. 

After  having  arranged  the  polynomial  with  reference  to  a,  extract 
the  square  root  of  25  a*:  this  gives  ba^,  which  is  placed  at  the  right 
of  the  polynomial.  Then  divide  the  second  term,  —  30a^6,  by  the 
double  of  5a^,  or  10 a^:  the  quotient  is  —Sab,  which  is  placed  at  the 
right  of  5  a^.  Hence  the  first  two  terms  of  the  root  are  5  a^  —  3  ab. 
Squaring  this  binomial,  it  becomes  25  a*  —  30  a^6 -f  9  a^i^^  which,  sub- 
tracted from  the  given  polynomial,  leaves  a  remainder  of  which  the 
first  term  is  AOa^b"^.  Dividing  this  first  term  by  10  a'^  (the  double  of  Sa^), 
the  quotient  is  +  4  6^ :  this  is  the  third  term  of  the  root,  and  is  written 
on  the  right  of  the  first  two  terms.  By  forming  the  double  product 
of  da^  —  Sab  by  46^,  squaring  46^,  and  taking  the  sum,  we  find  the 
polynomial  40a'^62  — 24a6^  +  16M,  which,  subtracted  from  the  first 
remainder,  gives  0.     Therefore  5  a^  —  3  o6  4-  4  6^  is  the  required  root. 

If  a  final  remainder  is  found  equal  to  0,  the  root  is  exact ;  if  not,  the 
root  is  only  approximate. 

2.  Find  the  square  root  of  a*  +  4  a^x  +  6  aV  +  4  ao;'  +  x*. 

Ans,  a^  +  2ax-t-x'^. 

3.  Find  the  square  root  of  a*  —  4  a^x  +  6  aV  —  4  ax^  +  x^. 

Ans.  a^  —  2ax-\-  x'^. 

4.  Find  the  square  root  of  4:X^-{'12c(^  +  5x* —  2x^ +1  x"^ 

-2:r  +  l.  Ans.  2a^  +  Sx'-x  +  l. 

5.  Find  the  square  root  of   9a* —  12 a'b -{- 280^1^ -IQab^ 

+  165*.  Ans.  Sa'-2ab  +  ^b\ 

6.  What   is    the    square   root   of   :r*— 4a:r^+ 4aV  —  4a;^ 

+  Sax  +  4:?  Ans.  x^-2ax  —  2. 

7.  What  is  the  square  root  of  9.x^ —  12x  +  6x^ +  y^ —ii/ 

+  4?  Ans.  3x  +  y-2. 

8.  What  is  the  square  root  of  y*  —  2yV -\- 2x^  —  27/^  +  1 

+  x*?  Ans.  y'-x'-l. 

9.  What  is  the  square  root  of  9a'b*  -  30  a'b'  +  25a'b'  ? 

.  Ans.  3  a^b'^  —  5  ab. 


BOOTS    OF    POLYNOMIALS.  201 

10.    Find   the    square    root    of    25a*Z>' -  40a'5V  +  76a'&V 

-  ASab'c^  +  36  bV  -  30  a*bc  +  24  a'bc'  -  36  a'bc'  +  9  aV. 

Ans.  5  d%  —  3  a'^c  —  4  a6c  +  6  bc^. 

150.  We  will  conclude  this  subject  with  the  following 
remarks :  — 

(1)  A  binomial  can  never  be  a  perfect  square,  since  we 
know  that  the  square  of  the  most  simple  polynomial  (viz.,  a 
binomial)  contains  three  distinct  parts,  which  cannot  expe- 
rience any  reduction  amongst  themselves.  Thus,  the  expression 
d^  +  b'^  is  not  a  perfect  square :  it  wants  the  term  =b  2ab,  in 
order  that  it  should  be  the  square  of  a±b. 

(2)  In  order  that  a  trinomial,  when  arranged,  may  be  a 
perfect  square,  its  two  extreme  terms  must  be  squares,  and 
the  middle  term  must  be  the  double  product  of  the  square 
roots  of  the  two  others.  Therefore,  to  obtain  the  square 
root  of  a  trinomial  when  it  is  a  perfect  square. 

Extract  the  roots  of  the  two  extreme  teo^ms,  and  give  these 
roots  the  same  or  contrary  signs,  according  as  the  middle  term 
is  positive  or  negative.  To  verify  it,  see  if  the  double  product 
of  the  tivo  roots  is  the  same  as  the  middle  terTu  of  the  trinomial. 

Thus,       9  a®  —  A&a^h^  +  64a^5*  is  a  perfect  square, 


since  V9a^  =  3 a'  and  V64 a^b'  =  -Sab'', 

and  also 

2xSa^X  —  Sab^  =  —  48a^Z>'  =  the  middle  term. 

r 

But  4a^  +  14a5  +  9  ^^  is  not  a  perfect  square  :  for  although 
4a'^  and  +95^  are  the  squares  of  2a  and  Sb,  yet  2  X  2a  X  3^ 
is  not  equal  to  14:  ab. 

(3)  In  the  series  of  operations  required  by  the  general 
rule,  when  the  first  term  of  one  of  the  remainders  is  not 


202  ELEMENTARY    ALGEBRA. 

exactly  divisible  by  twice  the  first  term  of  the  root,  we  may 
conclude  that  the  proposed  polynomial  is  not  a  perfect  square. 
This  is  an  evident  consequence  of  the  course  of  reasoning  by 
which  we  have  arrived  at  the  general  rule  for  extracting  the 
square  root. 

(4)  When  the  polynomial  is  not  a  perfect  square,  it  may 
sometimes  be  simplified  (see  §  139).  ' 

Take,  for  example,  the  expression  -y/a^h  -{-^aV  +  4a6^ 

The  quantity  under  the  radical  sign  is  not  a  perfect  square  ; 
but  it  can  be  put  under  the  form  ah{o^  +  4a5  +  4^^).  Now, 
the  factor  within  the  parenthesis  is  evidently  the  square  of 
a  +  25,  whence  w^e  may  conclude  that 


Va^5  +  4 a'h^  +  4 ah'  =-  (a  +  2  h)^ab. 


Take  also  the  expression  -sj'ld^b  —  4:ab'^  +  26*. 
V2  ^^5-"4  aF+"2^"^  =={a-h)  V2b. 

Cube  Root  of  Polynomials. 

150  a.  From  the  rules  for  forming  the  powers  of  binomials, 
we  have 

(a  +  by  =  a'  +  Sa'b  +  ^ab'  +  b\ 

Hence  the  cube  of  a  binomial  is  arranged  with  reference  to 
the  descending  powers  of  its  first  term,  and  is  equal  to  the 
cube  of  the  first  term  plus  three  times  the  square  of  the  first 
term  multiplied  by  the  second,  plus  other  terms. 

In  reversing  the  process,  to  find  the  cube  root  of  a  polyno- 
mial containing  not  more  than  four  terms,  we  must  arrange 
it  with  reference  to  one  of  its  letters ;  then  find  the  cube  root 
of  the  first  term  for  the  first  term  of  the  root,  and  subtract  its 
cube  from  the  polynomial.  The  first  term  of  the  remainder 
will  be  three  times  the  square  of  the  first  term  of  the  root 
multiplied  by  the  second ;  and  therefore,  to  find  the  second 


ROOT:^    OF    POLYNOMIALS.  203 

term  of  the  root,  divide  the  first  term  of  the  remainder  by- 
three  times  the  square  of  the  first  term  of  the  root. 

As  an  example,  extract  the  cube  root  of  a^  +  3  a^b  +  3  a6^  +  b^. 

a^  +  3a'b  +  Sab'  +  b'\a-^b 
a^  =  a^  \Sa^,  Divisor. 


3a=^^>  +  etc. 
(a  +  by  =  a'  +  Sa'b  +  Sab'  +  b' 

0     The  root  is  exact. 

Suppose  it  be  required  to  raise  a;  +  3/  +  c  to  the  third 
power.  We  may  write  it  in  the  foron  of  a  binomial,  thus : 
(a;  +  y)  +  c,  in  which  the  sum  of  x  and  3/  is  taken  for  the 
first  term.     Then,  as  before,  we  have 

[(^  +  y)  +  cJ--(x  +  yy  +  S(x  +  7/yc  +  B{x  +  y)c'  +  c\ 

Now,  the  cube  root  of  the  second  member  has  three  terms. 
The  first  two  terms,  x  and  y,  are  obtained  just  as  in  the  pre- 
ceding example.  When  the  cube  of  {x  +  y),  or  {x  +  3/)^,  is 
subtracted,  there  will  be  a  second  remainder  consisting  of 
3  (rr  +  yyc,  plus  other  terms.  The  first  term  of  this  remainder 
is  Sx'c;  that  is,  it  is  three  times  the  square  of  the  first  term 
of  the  root  multiplied  by  the  third  term  of  the  root ;  and 
hence,  to  find  the  third  term  of  the  root,  divide  the  first  term 
of  the  second  remainder  by  three  times  the  square  of  the  first 
term  of  the  root. 

To  illustrate  by  an  example,  extract  the  cube  root  of 

a;«  -  ^x'  +  Ibx'  -  20a^  +  I5x'  -6x+l, 

x^-6x'+15x'-20x'^+15x'-ex+l\x'^2x+l 
(x''~2xy  =  x^-6x'+12x*~   8x^  [3£*, Divisor. 

3  a;*- etc. 
(x'-2x+iy  -  x^--6x'+l5x'-20x'+15x'-6x+l 

0 


204  ELEMENTARY    ALGEBRA. 

In  this  example,  we  first  extract  the  cube  root  of  x^,  which 
gives  x^,  for  the  first  term  of  the  root.  Squaring  x"^,  and  mul- 
tiplying by  3,  we  obtain  the  divisor  3:r* :  this  is  contained  in 
the  second  term  —  2x  times.  Then  cubing  the  sum  of  these 
two  terms,  and  subtracting,  we  find  that  the  first  term  of  the 
remainder,  3  a;*,  contains  the  divisor  once.  Cubing  the  sum 
of  the  three  terms  and  subtracting,  we  find  a  remainder  0. 
Hence  a^  ~2x  +  \  is  the  exact  cube  root. 

Hence  we  have  the  following  rule  :  — 

Arrange  the  given  polynomial  with  reference  to  one  of  its 
letters,  and  extract  the  cube  root  of  the  first  term.  This  will  be 
the  first  term  of  the  root. 

Divide  the  second  term,  of  the  polynomial  by  three  times  the 
square  of  the  first  term  of  the  root.  The  quotient  will  be  the 
second  term  of  the  root. 

Subtract  the  cube  of  the  sum  of  the  two  terms  of  the  root  from 
the  given  polynomial,  and  divide  the  first  term  of  the  remainder 
by  three  times  the  square  of  the  first  term  of  the  root.  The 
quotient  will  be  the  third  term  of  the  root.  ^ 

Continue  this  operation  till  a  remainder  0  is  found,  or  until 
one  is  found  whose  first  term  is  not  divisible  by  three  times  the 
square  of  the  first  term  of  the  root. 

In  the  former  case  the  root  is  exact :  in  the  latter  case  the 
polynomial  is  an  imperfect  third  power. 

Exercises. 
Find  the  cube  roots  of  the  following  polynomials  :  — 

1.  8a;^-12.r^  +  6a:-l.  Ans.  2a; -1. 

2.  a^  +  lbx^+1bx+l2b.  Ans.  x  +  b. 

3.  x^  +  6a;^  -  40a;'  +  96a;  -  64.  Ans.  o;^  +  2a;  -  4. 

4.  8a;«  -  12a;^  +  30a;*  -  25a;«  +  30a;^  -  12a;  +  8. 

Ans.  2a;'^-a;-f  2. 


ROOTS    OF    POLYNOMIALS.  205 

5.  64 a'' -  288 a^+ 1080 a' -1458 a -729. 

Arts.  4a^  —  6  a  — 9. 

6.  l-6x  +  2\x'  -  44a;^  +  63a;*  -  54rt;^  +  27a;«. 

Ans.  1  — 207  +  30:". 

By  extracting  the  required  root  of  the  first  and  the  last 
terms,  two  terms  of  the  root  may  in  general  be  found,  from 
which  the  remaining  ones  may  often  be  determined  by  inspec- 
tion.    The  whole  root  may  then  be  verified  as  above. 


CHAPTER   VIII. 

EQUATIONS   OF  THE   SECOND   DEGREE. 

Equations  containing  One  Unknown  Quantity. 

151.  An  equation  of  the  second  degree  containing  but  one 
unknown  quantity  is  one  in  which  the  greatest  exponent  of 
the  unknown  quantity  is  equal  to  2.     Thus, 

x^  =  a,    ax^  -{-bx  —  Cy 

are  equations  of  the  second  degree. 

152.  Let  us  see  to  ^  what  form  every  equation  of  the  second 
degree  may  be  reduced. 

Take  any  equation  of  the  second  degree,  as 

^  ^       4  4      2 

Clearing  of  fractions,  and  performing  indicated  operations, 
we  have 

4:  +  Sx  +  4:x'  —  ox  -  4:0  =  20  -  X  +  2x\ 

Transposing  the  unknown  terms  to  the  first  member,  the 
known  terms  to  the  second,  and  arranging  with  reference  to 
the  powers  of  x,  we  have 

4.r'  -  2^'  +  8a;  ~  3a:  +  a;  =  20  +  40  -  4. 

By  reducing,  2a;^  +  6a;  =  56. 

Dividing  by  the  coefficient  of  a;^  we  have 

.r»  +  8a':^28. 
200 


EQUATIONS    OF    THE    SECOND    DEGREE.  207 

If  we  denote  the  coefficient  of  x  by  2j9,  and  the  second 
member  by  q,  we  have 

x^  +  2px  =  q. 
This  is  called  the  reduced  equation. 

153.  When  the  reduced  equation  is  of  this  form,  it  contains 
three  terms,  and  is  called  a  complete  equation.  The  terms 
are,  — 

First  Term.  —  The  second  power  of  the  unknown  quan- 
tity, with  a  plus  sign. 

Second  Term.  —  The  first  power  of  the  unknown  quantity, 
with  a  coefficient. 

Third  Term.  —  A  known  term  in  the  second  member. 

Every  equation  of  the  second  degree  may  be  reduced  to 
this  form  by  the  following  rule  :  — 

Clear  the  equation  of  fractions,  and  perform  all  the  indicated 
operations. 

Transpose  all  the  U7iknown  terms  to  the  first  member,  and 
all  the  known  terms  to  the  second  unember. 

Reduce  all  the  terms  containing  the  square  of  the  unknown 
quantity  to  a  single  term,  one  factor  of  which  is  the  square  of 
the  unknown  quantity ;  reduce  also  all  the  terms  containing 
the  first  power  of  the  unknoiun  quantity  to  a  single  term. 

Divide  both  viembers  of  the  resulting  equation  by  the  coeffi- 
cient of  the  square  of  the  unknown  quantity. 

154.  A  root  of  an  equation  is  sucK  a  value  of  the  unknown 

quantity  as,  being  substituted  for  it,  will  satisfy  the  equation ; 
that  is,  make  the  two  members  equal* 

The  solution  of  an  equation  is  the  operation  of  finding  its 
roots. 


208  ELEMENTARY    ALGEBRA. 


Incomplete  Equations. 


155.  It  may  happen  that  2p,  the  coefficient  of  the  first 
power  of  X,  in  the  equation  x  +  2px  —  q,  is  equal  to  0.  In 
this  case  the  first  power  of  x  will  disappear,  and  the  equation 
will  take  the  form 

x-'  =  q  (1) 

This  is  called  an  incomplete  equation.     Hence 

An   incomplete   equation,  when   reduced,  contains   but  two 

terras,  —  the  square  of  the  unknown  quantity,  and  a  known 

term. 

156.  Extracting  the  square  root  of  both  members  of  Equa- 
tion (1),  we  have 

a;  -  ±  V^  (2) 

Hence,  for  the  solution  of  incomplete  equations, 
Reduce  the  equation  to  the  form  o^  =  q. 
Then  extract  the  square  root  of  both  members. 

Note.  —  There  will  be  two  roots,  numerically  equal,  but  having  con- 
trary signs.     Denoting  the  first  by  a;',  and  the  second  by  x^\  we  have 

x'  =  +  y/q,  and  x^^  =  —  V^. 

Verification.  —  Substituting  +  y/q  or  —  y/q  for  x,  in  Equation 
(1),  we  have  (+  y/qf  =  gr,  and  (-  V^)^  =  q.  Hence  both  satisfy  the 
equation  :  they  are  therefore  roots  (§  154). 

Exercises. 
1.    Whatare  the  values  of  a;  in  the  equation  3a;''+8  =  5a;'  — 10? 
By  transposing,  3  x*  -  5  rc^  =  -  10  -  8. 

Reducing,  -  2  a;"  =  -  18. 

Dividing  by  —  2,  a;^  =  9. 

Extracting  square  root,  a;  =  i  V9  =»  +  3  and  —  3. 

Hence  x'  -  +  3,  and  x^^  »  -  3. 


EQUATIONS    OF    THE    SECOND    DEGREE.  209 

2.  What  are  the  roots  of  the  equation  3  :i;^  +  6  ==  4  a:^  —  10  ? 

Ans.  :r'  ==  +  4,  a;"  =  —  i. 

1  x^ 

3.  What  are  the  roots  of  the  equation  -x^  —  S  =  — hlO? 

o  y 

Ans.  x'  =  +  9,  07"  =  — 9. 

4.  What  are  the  roots  of  the  equation  4:r^  +  13  —  2a;^  =  45  ? 

Ans.  x'  =  +  4c,  x''  =  —  4:. 

6.    What  are  the  roots  of  the  equation  Gx"^  —  7  =  Sx^  -}-  b? 

Ans.  x'  =  +  2,  x"  =  —  2, 

6.  What  are  the  roots  of  the  equation  S  +  5x''=—  +  4:X^  +  2S? 

5 

Ans.  x'  =  -\-5,  x''  —  —  5. 

7.  What   are  the  roots  of  the  equation  — '^-^ — 

-117-5:r^?  ^  ^ 

Ans.  x'  ==-{-  5,  x^'  =  —  5. 

8.  What  are  the  roots  of  the  equation  x'^  +  ab  =  5x'^? 

Ans.  x'  =  -{--  -Vab,  x"  = Va^. 


9.    What  are  the  roots  of  the  equation  .r  Va  -{-  x"^  =  b  -}-  x^  ? 

Ans.  x'  =  —  ,  x'^  = • 

■\^a  —  2b  Va-26 


Problems  for  Solution. 

1.    What  number  is  that  which  being  multiplied  by  itself 
the  product  will  be  144  ? 

Let  X  =  the  number. 

.  Then  xx  x  =  x^  =  144. 

It  is  plain  that  the  value  of  x  will  be  found  by  extracting  the  square 
root  of  both  members  of  the  equation  ;  that  is, 

Vx^  =  \/l44  ;    that  is,  x  =  12. 

D.  N.  E.  A.  —  14. 


210  ELEMENTARY    ALGEBKA. 

2.  A  person,  being  asked  liow  mucli  money  he  had,  said, 
"  If  the  number  of  dollars  be  squared  and  6  be  added,  the  sum 
will  be  42."     How  much  had  he  ?  Arts.  $Q. 

Let  X  =  the  number  of  dollars. 

Then,  by  the  conditions,     a;^  +  6  =  42. 
Hence  a;^  _  42  -  6  =  36, 

and  x  =  6. 

3.  A  grocer,  being  asked  how  much  sugar  he  had  sold  to 
a  person,  answered,  "  If  the  square  of  the  number  of  pounds 
be  multiplied  by  7,  the  product  will  be  1575."  How  many 
pounds  had  he  sold  ?  Ans.  15. 

Let  X  =  the  number  of  pounds. 

Then,  by  the  conditions  of  the  question, 

7a;2  =  1575. 
Hence  '  x^==225, 

and  a;  =  15. 

4.  A  person,  being  asked  his  age,  said,  "  If  from  the  square 
of  my  age  in  years  you  take  192  years,  the  remainder  will  be 
the  square  of  half  my  age."     What  was  his  age  ? 

Ans.  16  years. 

Let  X  =  the  number  of  years  in  his  age. 

Then,  by  the  conditions  of  the  question, 


Clearing 

I  of  fractions, 

•4x2-768  =  0:2. 

Hence 

4ar'-a;2=768, 

and 

3a;2  =  768. 
x^  -  256. 
X  «  16. 

EQUATIONS    OF   THE   SECOND    DEGREE.  211 

6.  What  number  is  that  whose  eighth  part  multiplied  by 
its  fifth  part,  and  the  product  divided  by  4,  will  give  a  quo- 
tient equal  to  40  ?  Ans,  80. 


Let 

X  ^ 

=  the  number. 

By  the  conditions  of  the 

question, 

1 

(1-4-) 

-4  = 

=  40. 

Hence 

160 

=  40. 

Clearing  of  fractions, 

x^  = 

=  6400. 

x  = 

=  80. 

6.  Find  a  number  such  that  one  third  of  it  multiplied  by 
one  fourth  shall  be  equal  to  108.  Ans.  36. 

7.  What  number  is  that  whose  sixth  part  multiplied  by  its 
fifth  part,  and  the  product  divided  by  10,  will  give  a  quotient 
equal  to  3  ?  Ans.  30. 

8.  What  number  is  that  whose  square  plus  18  will  be  equal 
to  half  the  square  plus  30-|-  ?  Ans.  5. 

9.  What  numbers  are  those  which  are  to  each  other  as  1 
to  2,  and  the  difference  of  whose  squares  is  equal  to  75  ? 

Ans.  5  and  10. 

Let  X  =»=  the  less  number. 

Then  2x=  the  greater. 

Then,  by  the  conditions  of  the  question, 

4a;2-a;2  =  75. 
Hence  3a;2  =  75. 

Dividing  by  3,  x^  =  25,  and  x  =  5, 

and  2j?»10. 


212  ELEMENTARY    ALGEBRA. 

10.    What  two  numbers  are  those  which  are  to  each  other 
as  5  to  6,  and  the  difference  of  whose  squares  is  44  ? 

Ans.  10  and  12. 


Let 

X  = 

=  the  greater 

number. 

Then 

^x- 
6 

=  the  less. 

By  the  conditions  of  the 

problem, 

x^- 

36 

=  44. 

Clearing 

of  fractions, 

•66  x^ - 

-25x^  = 

=  1584. 

Hence 

lla;--'  = 

=  1584, 

and 

a;2  = 

=  144. 

Hence 

x  = 

=  12, 

and 

6 

=  10. 

11.  What  two  numbers  are  those  which  are  to  each  other 
as  3  to  4,  and  the  difference  of  whose  squares  is  28  ? 

Ans.  6  and  8. 

12.  What  two  numbers  are  those  which  are  to  each  other 
as  5  to  11,  and  the  sum  of  whose  squares  is  584? 

Ans.  10  and  22. 

13.  A  says  to  B,  "  My  son's  age  is  one  quarter  of  yours, 
and  the  difference  between  the  squares  of  the  numbers  repre- 
senting their  ages  is  240."     What  were  their  ages  ? 

Ans.  Elder,  16  ;  younger,  4. 

Equations  containing  Two  Unknown  Quantities. 

157.    When  there  are  two  or  more  unknown  quantities, 

Eliminate  one  of  the  unknown  quantities  hy  §  113. 

Then  extract  the  square  root  of  both  members  of  the  equation. 


EQUATIONS  OF  THE  SECOND  DEGREE.         213 

Problems  for  Solution. 

1.  There  is  a  room  of  such  dimensions  that  the  difference 
of  the  sides  multiplied  by  the  less  is  equal  to  36,  and  the 
product  of  the  sides  is  equal  to  360.     What  are  the  sides  ? 

Ans.  a;  =  18,  7/  =  20. 

Let  X  =  the  length  of  the  less  side, 

and  3/  =  the  length  of  the  greater. 

Then,  by  the  1st  condition,     (y  —  x)  x  =  36  ; 
and  by  the  2d,  xy  =  360. 

From  the  first  equation  we  have 

xy  —  x"^  =  36; 

and  by  subtraction,  x^  =  324. 

Hence  x  =  V324  =  18, 

and  y  =  ^==20. 

•^18 

2.  A  merchant  sells  two  pieces  of  muslin,  which  together 
measure  12  yards.  He  received  for  each  piece  just  as  many 
dollars  per  yard  as  the  piece  contained  yards.  Now,  he  gets 
four  times  as  much  for  one  piece  as  for  the  other.  How  many 
yards  in  each  piece  ?  Ans.  8  and  4. 

Let  x  =  the  number  of  yards  in  the  longer  piece, 

and  y  =  the  number  of  yards  in  the  shorter  piece. 

Then,  by  the  conditions  of  the  question, 
jc  -f  3/  =  12. 
xXx  =  x^  =  what  he  got  for  the  longer  piece. 
y  Xy  ^y"^  =  what  he  got  for  the  shorter. 
x^  =  4y^,  by  the  2d  condition, 
a;  =  2 1/,  by  extracting  the  square  root. 
Substituting  this  value  of  a;  in  the  first  equation,  we  have 
3/ +  22/ =  12. 
and  consequently  y  =  4, 

and  X  =  8.  ,  . . 


214  ELEMENTARY    ALGEBRA. 

3.  What  two  numbers  are  those  whose  product  is  30,  and 
the  quotient  of  the  greater  by  the  less,  3-^  ?       Ans.  10  and  3. 

4.  The  product  of  two  numbers  is  a,  and  their  quotient  b. 

What  are  the  numbers  ?  ,— -  C 

Ans.  -Vab  and  -i/-- 

5.  The  sum  of  the  squares  of  two  numbers  is  117,  and  the 
difference  of  their  squares  45.     What  are  the  numbers  ? 

Ans.  9  and  6. 

6.  The  sum  of  the  squares  of  two  numbers  is  a,  and  the 
difference  of  their  squares  is  h.     What  are  the  numbers  ? 


^^s-  \^4—    and    ^~ 


2  ^2 

7.  What  two  numbers  are  those  which  are  to  each  other 
as  3  to  4,  and  the  sum  of  whose  squares  is  225  ? 

Ans.  9  and  12. 

8.  What  two  numbers  are  those  which  are  to  each  other 
as  m  to  w,  and  the  sum  of  whose  squares  is  equal  to  a^  ? 

Ans.  — z=z=zz=:    and 


Vm""^  +  n^  -yjm^  +  n^ 

9.  What  two  numbers  are  those  which  are  to  each  other 
as  1  to  2,  and  the  difference  of  whose  squares  is  75  ? 

Ans.  5  and  10. 

10.  What  two  numbers  are  those  which  are  to  each  other 

as  7?i  to  72,  and  the  difference  of  whose  squares  is  equal  to  li^  ? 

A  nib  J  nb 

Ans.  — =::=rr   and 


^ni^  —  n?  Vm'^  —  n^ 

11.  A  certain  sum  of  money  is  placed  at  interest  for  six 
months,  at  8  per  cent  per  annum.  Now,  if  the  sum  put  at 
interest  be  multiplied  by  the  number  expressing  the  interest, 
the  product  will  be  $562500.  What  is  the  principal  at 
interest?  Ans.  $3750. 


EQUATIONS    OF   THE   SECOND    DEGREE.  215 

12.  A  person  distributes  a  sum  of  money  between  a  num- 
ber of  women  and  boys.  The  number  of  women  is  to  the 
number  of  boys  as  3  to  4.  Now,  the  boys  receive  one  half 
as  many  dollars  as  there  are  persons ;  and  the  women,  twice 
as  many  dollars  as  there  are  boys ;  and  together  they  receive 
$188.     How  many  women  were  there,  and  how  many  boys? 

Ans.  36  women,  48  boys. 


Complete  Equations. 

158.   The  reduced  form  of  the  complete  equation  (§  153)  is 

x^  +  2px  —  q. 

Comparing  the  first  member  of  this  equation  with  the 
square  of  a  binomial  (§  54),  we  see  that  it  needs  but  the 
square  of  half  the  coefficient  of  x  to  make  it  a  perfect  square. 
Adding  p^  to  both  members  (§  102,  Axiom  1),  we  have 

x'^  +  2px  +p'^  —  q  +  p^. 

Then,  extracting  the  square  root  of  both  members  (Axiom  5), 
we  have 


x+p  =  ^-\/q+p\ 
Transposing  p  to  the  second  member,  we  have 


X  —  —p  ifVq  +/>^ 

Hence  there  are  two  roots,  —  one  corresponding  to  the  plus 
sign  of  the  radical,  and  the  other  to  the  minus  sign.  De- 
noting these  roots  by  x^  and  x^\  we  have 


x^  ~  —  p +-\/ q  +  p\  and  a:"  =  —  jo  —  V^' +j9^ 

The  root  denoted  by  re'  is  called  the  first  root ;  that  denpte^ 
by  x^^  is  called  the  second  root. 


216  ELEMENTARY   ALGEBRA. 

159.  Oompleting  the  square  is  the  operation  of  squaring  half 
the  coefficient  of  x,  and  adding  the  result  to  both  members 
of  the  equation.  For  the  solution  of  every  complete  equation 
of  the  second  degree,  we  have  the  following  rule :  — 

Meduce  the  equation  to  the  form  x^  +  ^jpx  =  q. 

Take  half  the  coefficient  of  the  second  term,  square  it,  and 
add  the  result  to  both  members  of  the  equation. 

Then  extract  the  square  root  of  both  rtiembers,  after  which 
transpose  the  known  term  to  the  second  member. 

Note.  —  Although  in  the  beginning  the  student  should  complete  the 
square  and  then  extract  the  square  root,  yet  he  should  be  able  in  all 
cases  to  write  the  roots  immediately  by  the  following  (see  §  158)  rule :  — 

The  first  root  is  equal  to  half  the  coefficient  of  the  second 
term  of  the  reduced  equation^  taken  with  a  contrary  sign,  plus 
the  square  root  of  the  second  member  increased  by  the  square 
of  half  the  coefficient  of  the  second  term. 

The  second  root  is  equal  to  half  the  coefficient  of  the  second 
term  of  the  reduced  equation,  taken  with  a  contrary  sign,  minus 
the  square  root  of  the  second  member  increased  by  the  square 
of  half  the  coefficient  of  the  second  term, 

160.  We  will  now  show  that  the  complete  equation  of  the 
second  degree  will  take  four  forms,  dependent  on  the  signs 
of  2p  and  q. 

Let  us  suppose  2p  to  be  positive,  and  q  positive :  we  shall 
then  have 

x''  +  2px  -=  q  (1) 

Let  us  suppose  2p  to  be  negative,  and  q  positive :  we  shall 
then  have 

x^  —  2px  =  q  (2) 

Let  us  suppose  2p  to  be  positive,  and  q  negative :  we  shall 
then  have 

x''  +  2px-=-q  (3) 


EQUATIONS   OF    THE   SECOND   DEGREE.  217 

Let  US  suppose  2p  to  be  negative,  and  q  negative :  we  shall 
then  have 

x^  —  2px  ~  —  q  (4) 

As  these  are  all  the  combinations  of  signs  that  can  take 
place  between  2p  and  q^  we  conclude  that  every  complete 
equation  of  the  second  degree  will  be  reduced  to  one  or  the 
other  of  these  four  forms  :  — 


x"  +  2px  -=  +  q, 

1st  form. 

x^  -  2px  =  +  qy 

2d  form. 

x"  +  2px  =  -q, 

3d  form. 

x^  —  2px  ~  —  q, 

4th  form. 

Exercises. 

First  Form. 

1.   What  are  the  values  of  x  in  the  equation  2a;^  +  8  a;  =  64  ? 

If  we  first  divide  by  the  coefficient  2,  we  obtain 
x^  -\-  4  a;  =  32. 

Completing  the  square,       a;^  -f  4  a;  +  4  =  32  +  4  =  36. 

Extracting  the  root,      a;  +  2  =  ±  V36  =  +.6  and  —  6. 

Hence  a;''=  —  2-|-6  =  +4, 

and  a;^^  =  -  2  -  6  =  -  8. 

Hence,  in  this  form,  the  smaller  root,  numerically,  is  positive,  and 
the  larger  negative. 

Verification.      If  we  take  the  positive  value  (viz.,  a;'  =  +  4),  the 
equation  a;^  +  4  a;  =  32 

gives  42  +  4  X  4  =  32 ; 

and  if  we  take  the  negative  value  of  x  (viz.,  x^^  =  —  8),  the  equation 

a;2  +  4a;  =  32 
gives  (-8)2  +  4(-8)  =  64-32=32; 

from  which  we  see  that  either  of  the  values  of  x  (viz., 

a;^  =  +  4,  or  x^^  =  -  8) 
will  satisfy  the  equation. 


218  ELEMENTARY    ALGEBRA. 

2.  What  are  the  values  of  :r  in  the  equation 

3^;^  +  12a;  -  19  =  -  a;^  -  12^:  +  89  ? 

Transposing  the  terms,  we  have 

3x^  +  x^  +  \2x  +  12a;  =  89  +  19. 
Reducing,  4  a;^  +  24  a;  =  108. 

Dividing  by  the  coefficient  of  x"^, 

a;2  + 6a;  =  27. 
Completing  the  square,      a;^  +  6  a;  +  9  =  36. 

Extracting  the  square  root, 

a;  +  3  -■=  ±  \/36  =+  6  and  -  6. 

Hence  a;/=  +  6-3-  +  3, 

and  a;^^  =  -  6  -  3  =  -  9. 

Verification.      If  we  take  the  plus  root,  the  equation 
a;2  + 6a;  =  27 
gives  (3)2 +  6  (3)  =  27. 

If  we  take  the  negative  root,   a;^  +  6  a:  =  27 
gives  (-  9)2  +  6  (-  9)  =  81  -  54  =  27. 

3.  What  are  the  values  of  x  in  the  equation 

5 
Clearing  of  fractions,  we  have 

5x2  _  50a;  +  75  =  a;2  -  170x  +  775. 
Transposing  and  reducing,  we  obtain, 

4a;2  + 120  a;  =  700. 
Dividing  by  the  coefficient  of  a;^,  we  have 

a;^  +  30a;-»175. 
Completing  the  square,  x«  +  30.t  +  225  =  400. 


EQUATIONS    OF    THE    SECOND    DEGREE.  219 

Extracting  the  square  root, 

a;  +  15  =  ±  ViUo  =  +  20  and  -  20. 
Hence  a;''  =  -f  5,  and  x^^  =  —  35. 

Verification.      For  the  plus  value  of  x,  the  equation  ^ 
a;2  +  30a;=175 
gives  (5)2  +  30  X  5  =  25  +  150  =  175. 

For  the  negative  value  of  x,  we  have 

(_  35)2  +  30  (_  35) «  1225  -  1050  =  175. 

4.    What  are  the  values  of  x  in  the  equation 

-x^ —-x^-^S —-X  —  x^ -\ ? 

6         2        4  3  12 

Clearing  of  fractions,  we  have 

10^2  -6x4-9  =  96 -8a:-  12.r2  +  273. 

Transposing  and  reducing,  22a;2  +  2  a:  =  360. 

Dividing  both  members  by  22, 

2,2         360 

x^  -h  —  x  = 

22         22 

Add  ( —  )   to  both  members,  and  the  equation  becomes 

22        V22;       22       V22y 
Whence,  by  extracting  the  square  root, 


x-h 
22 


12  ^  22       \22y 


22      >22       V22; 

and  ,...^l_J^WfJ. 

22      ^  22       \22) 

It  remains  to  perform  the  numerical  operations.     In  the  first  place, 

360     /  j_y 
v^      ;  22     [22) 

must  be  reduced  to  a  single  number,  having  (22)^  for  its  denominator. 

\T     .        ^^(XY     360  X  22  -f- 1  ^  7921 
'  '^^^'        22     A  22/  "^         (22)2         "'(22)«" 


220  ELEMENTARY   ALGEBRA. 

Extracting  the  square  root  of  7921,  we  find  it  to  be  89. 


••• 

*Vf 

*{ 

22/ 

22 

Consequently 

the  pli 

IS  value  of  x  is 

f 

a;^  =  - 

1 
'22 

+  §?  = 

22 

22       ' 

and  the  negative 

value 

is 

x'^  =  - 

1 
"22 

89 
22 

45. 
11' 

that  is,  one  of  the  two  values  of  x  which  will  satisfy  the  proposed  equa- 
tion is  a  positive  whole  number,  and  the  other  a  negative  fraction. 

Note.  —  Let  the  pupil  be  exercised  in  writing  the  roots  in  the  last 
five  and  in  the  following  examples  without  completing  the  square. 

5.  What  are  the  values  of  x  in  the  equation 

3:r^  +  2a;  -  9  =:  76?  Ans.  j  ^'^  ^'  _ 

6.  What  are  the  values  of  x  in  the  equation 

7.  What  are  the  values  of  x  in  the  equation 

8.  What  are  the  values  of  x  in  the  equation 

9.  What  are  the  values  of  x  in  the  equation 

X     I    X        X  X     ,    ±o  o  A  \  X    —  i.. 

2  +  4  =  1-10+20^  ^."^•ix"=-2i. 


EQUATIONS    OF   THE   SECOND    DEGREE.  221 

Second  Foem. 

1.  What  are  the  values  of  x  in  the  equation 

a;^- 8a; +  10  =  19? 
Transposing,  a;^  _  3  j.  ,^  19  _  10  =  9. 

Completing  the  square, 

a;2_  8a; +  16  =  9  +  16  =  25. 
Extracting  the  root, 

a;  —  4  =  db  V25  =  +  5  or  —  5. 
Hence  o;^  =  4  +  5  =  9, 

and  x^^=4:  —  5  =  —  l. 

That  is,  in  this  form,  the  larger  root,  numerically,  is  positive,  and 
the  lesser  negative. 

Verification.     If  we  take  the  positive  value  of  x,  the  equation 

a;2  _  8  a;  =  9 

gives  (9)2-8x9  =  81-72  =  9. 

If  we  take  the  negative  value,  the  equation 

a;2_  8a;  =  9 

gives  (-l)2-8(-l)  =  l  +  8  =  9, 

from  which  we  see  that  both  roots  alike  satisfy  the  equation. 

2.  What  are  the  values  of  x  in  the  equation 

f  +  |-15  =  |  +  :.-14i? 

Clearing  of  fractions,  we  have 

6 x2  +  4a;  -  180  =  3  a;2  +  12a;  -  177. 
Transposing  and  reducing, 

3a;2-8a;  =  3. 
Dividing  by  the  coefficient  of  a;^,  we  obtain 

a;2 a;  =  1. 

3 


222  ELEMENTARY    ALGEBRA. 

Completing  the  square,  we  have 


3         9  9       9 


Extracting  the  square  root, 

4  /25       ,  5        ,      5 

Hence  a;/  =  i  +  5=+3,  and  a;^^  =  i-5  =  _i 

3      3  3      3         3 

Verification.      For  the  positive  root  of  x,  the  equation 

x^  —  a;  =  1 
3 

gives  32-^x3  =  9-8  =  1. 


For  the  negative  root,  the  equation 


x'^  —  a;  =  1 
3 


.  /     IV      8^      1      1   ,  8      1 
gives  ^__)_-x--  =  -+-  =  l. 

3.    What  are  the  values  of  x  in  the  equation     - 

2      3       ^ 
Clearing  of  fractions,  and  dividing  by  the  coefficient  of  x,  we  have 


Completing  the  square,  we  have 

2     2     ,1      11,1      49 
3        9      ^      9      36 

Extracting  the  square  root,  we  have 

Hence       .c^  =  1  +  -  =  -=  U.  and  x^^  -  1-^  =  ~ -• 
3      6     6^  366 


EQUATIONS    OF    THE    SECOND    DEGREE.  223 

Verification.       If  we  take  the  positive  root  of  x,  the  equation 

2 
7?-  —  -x=  \\ 

gives  (l|f-|xli  =  2i-l  =  li. 

If  we  take  the  negative  root,  the  equation 
x'-\x=^\\ 
{     ny     2         5     25     10     45     Ti 

4.    What  are  the  values  of  x  in  the  equation 
4:a:'--2x''  +  2ax=lSab-lSb'? 

By  transposing,  changing  the  signs,  and  dividing  by  2,  the  equation 
becomes  x"^  —  ax-  =  2  a^  —  9  a6  +  9  b'^. 

Completing  the  square,      x"^  —  ax  +  —-  =  '--    —9ab-r9  b'^. 

4        4 


Extracting  the  square  root,  x  =  -± 'V'— 9 a6  +  9 6^ 

JNow,  the  square  root  of  — ^  —  9ab  +  9b'^  is  evidently  —^  —  36. 
4  2 

.*.  re  =  -  ± oo],  and    < 

2     V  2  y  la;^^:=-    a  +  36. 

What  will  be  the  numerical  values  of  x,  if  we  suppose  a  =  6, 
and  5  =  1 ? 

5.  What  are  the  values  of  x  in  the  equation 

I  X  -  4:  -  X'  +  2X  --  X'  ■=  A5  -  Sx'  +  4:X? 

o  o 

^^s.    j  ^'  ^      I'l^  ]  to  within  0.01. 

6.  What  are  the  values  of  a;  in  the  equation 

Sx'~-Ux  +  10--=2x  +  M?  Ans.    {"^I^^^^* 


2:24  ELEMENTARY    ALGEBRA. 

7.    What  are  the  values  of  x  in  the  equation 


X' 


--30  +  :r-2:r-22? 
4 


Ans 


=  8. 


-4. 


8.    What  are  the  values  of  x  in  the  equation 


a;2_3^_|_|.^9^  +  13i.? 


9.    What  are  the  values  of  x  in  the  equation 

^ax-x^^-^ah-h^-t  Ans.    [^[r"^^^^' 

Lx"  =  —  b, 

10.    What  are  the  values  of  x  in  the  equation 
a'  +  b-'~2hx  +  x'  =  '^^'i 


Ans.  < 


-  {bn  +  -\/d^m^  +  ^^'^^^  —  o^n^). 


x'^  =  —^ — "2  {bn  —  y/o^m^  +  b^m^  —  aV). 


Third  Form. 

1.    What  are  the  values  of  x  in  the  equation  a;^  +  4a;  =  —  3? 
Completing  the  square,  we  have 

a;2  +  4a;  +  4  =  -3  +  4  =  l. 
Extracting  the  square  root, 

a;  +  2  =  ±\/i  =  +  l,  and  -  1. 
Hence  a;''  =  —  2  +  1  =  —  1; 

and  rc^^  =  -  2  -  1  =  -  3  ; 

that  is,  in  this  form  both  the  roots  are  negative. 

Verification.      If  we  take  the  first  negative  value,  the  equation 

gives  (-  1)2  +  4(-  1)  =  1  -  4  =  -  3. 


EQUATIONS   OF   THE   SECOND   DEGREE.  225 

If  we  take  the  second  value,  the  equation 

gives  (-  3)2  +  4(-  3)  =  9  -  12  =  -  3. 

Hence  both  values  of  x  satisfy  the  given  equation. 

2.  What  are  the  values  of  x  in  the  equation 

2  2 

Transposing  and  reducing,  we  have 

-.T2-lla;  =  28. 
Dividing  by  —  1,  the  coefficient  of  a;',  we  have 

a;2^11a;  =  -28. 
Completing  the  square, 

a;2  + 11  a; +  30,25  =  2.25. 
Hence  a;  +  5.5  =  ±  V2.25  =  +  1.5  and  —  1.5. 

Consequently         aj^  =  —  4,  and  x^^  =  —  7. 

3.  What  are  the  values  of  x  in  the  equation 

4.  What  are  the  values  of  x  in  the  equation 

5.  What  are  the  values  of  x  in  the  equation 

6.  WEat  are  the  values  of  x  in  the  equation 

_.._4-|.  =  ^+24.  +  2?        ^„^ 


D.  N.  K.  A.  —  15. 


x' 

= 

1 

3' 

x" 

= 

-4. 

x' 

= 

1 

5' 

—  2. 

x" 

= 

x' 

= 

1 
4 

x" 

= 

-8. 

8. 
10. 


226  ELEMENTARY    ALGEBRA. 

7.  What  are  the  values  of  x  in  the  equation 

-.r^+7a;  +  20=--a;^-lla;-60?   Am.   \^\~' 

8.  What  are  the  values  of  x  in  the  equation 

5a;'^_^  +  l:^_9i^_1^2_l^  \x'  =-\ 

6  2  ^62  ^^^s.   ^j  8 

I  a;"  =  -  8. 

9.  W^hat  are  the  values  of  x  in  the  equation 

[  a;"  -  -  10. 

10.  What  are  the  values  of  x  in  the  equation 

x  —  x^—Z  =  ^x-^\'l  Arts. 

11.  What  are  the  values  of  x  in  the  equation 

a;'  +  4:r-90  =  -93?  Arts. 

Fourth  Foem. 


(  rr'   -  -  1 
U"  =  -4 


4. 


fa;'  --1. 
1  a;"  -  -  3. 


1.    What  are  the  values  of  x  in  the  equation  x^  —'^x  —  ■ 
Completing  the  square,  we  have 

a:2_8a;  +  16  =  -7  +  16  =  9. 
Extracting  the  square  root, 

a;-4  =  ±V9=  +  3  and  -3. 
Hence  a;^  =  +  7,  and  a;^''  =  +  1 ; 

that  is,  in  this  form  both  the  roots  are  positive. 

Verification.     If  we  take  the  greater  root,  the  equation 
a;2_8a;  =  -7 
gives  72  -  8  X  7  =  49  -  56  =  -  7. 

If  we  take  the  lesser,  the  equation 

j,2_8.T  =  -7 

gives  P-8xl=    1-    8  =  -7. 

Hence  both  of  tho  roots  will  satisfy  the  equation. 


EQUATIONS   OF   THE   SECOND    DEGREE.  227 

2.  What  are  the  values  of  x  in  the  equation 

-  l^:r^  +  3:r  -  10  -  l|;r^  -  18a; +  — ? 

Clearing  of  fractions,  we  have 

-  3x2  -r  6ic  -20  =  3a;2  -  36a;  +  40. 
Collecting  the  similar  termp, 

-6a;2  +  42a;  =  60. 
Dividing  by  the  coefficient  of  x^,  which  is  —  6,  we  have 

x'-lx^-lO. 
Completing  the  square,  we  have 

a:2- 7  a: +  12.25  =  2.25. 
P^xtracting  the  square  root  of  both  members, 

a; —  3.5  =  ±\/2.25  =  +  1.5  and  —1.5. 
Hence        0:^-3.5  +  1.5  =  5,  and  a;^^=  3.5  -  1.5  =  2. 
Verification.     If  we  take  the  greater  root,  the  equation 
x'-lx^-lO 
gives  52  -  7  X  5  =  25  -  35  =  -  10. 

If  we  take  the  lesser  root,  the  equation 

x^  —  1  x  =  —  \0 
gives  22  -  7  X  2  =  4  -  14  =  -  10. 

3.  What  are  the  values  of  x  in  the  equation 

-  3  a;  +  2  a;^  +  1  =  1 7|  :r  -  2  a;^  -  3  ? 
Transposing  and  collecting  the  terms,  we  have 

4a;2-20fa;  =  -4. 
Dividing  by  the  coefficient  of  x^,  we  have 

.r2-5|a;  =  -l. 
Completing  the  square,  we  obtain 


228 


ELEMENTARY   ALGEBRA. 


Extracting  the  root, 


a:^-2l 


■■±M 


/1M=  +  L2and    -12. 
25  5  5 


Hence       a;^=  2f  +  ^  =  5,  and  a:^/=  2f  -  ?|  =  1- 
5  5       5 

Verification.     If  we  take  the  greater  root,  the  equation 


gives  52  -  5J  X  5  =  25  -  26  =  -  1. 

If  we  take  the  lesser  root,  the  equation 

fiy     .1^1       1       26         1 
g,ves  (5)-'^^5==25-25  =  -'- 

4.    What  are  the  values  of  x  in  the  equation 


1 


1 


Sx  +  -  =  -^x^  +  ^x-^? 


What  are  the  values  of  x  in  the  equation 


Ayis. 


6.    What  are  the  values  of  x  in  the  equation 
^20        40  20    ^40 


Ans. 


7.  What  are  the  values  of  x  in  the  equation 

x'-10-^x=^-l? 

8.  What  are  the  values  of  x  in  the  equation 


Ans. 


17 


2x' 


_27a:  +  ^  +  100  =  ^  +  12a:~26?  Ans. 
5  5 


X    = 
x^'  = 


3. 

.1 
'4* 


_1 

■7' 

_1 
'4* 

_1 
"5* 

10. 

10' 

=  7. 
=  6. 


EQUATIONS    OF    THE    SECOND    DEGREE.  229 

9.    What  are  the  values  of  x  in  the  equation 

3  3  \x"  =  l. 

10.    What  are  the  values  of  x  in  the  equation 


I  10 


Properties  of  Equations  of  the  Second  Degree. 

Mrsi  Property. 

161.  We  have  seen  (§  153)  that  every  complete  equation  of 
the  second  degree  may  be  reduced  to  the  form 

x'  +  2px^q  (1) 

Completing  the  square,  we  have 

x^  +  2px  -i-p'^  =  q+p^' 
Transposing  q  +  p"^  to  the  first  member, 

x'  +  2px+p'-(q+p')  =  0  (2) 

Now,  since  x^  +  2px  -{-p^  is  the  square  of  x+p,  and  q-\-p^ 
the  square  of  -\/q  -\-p'\  we  may  regard  the  first  member  as 
the  difference  between  two  squares.  Factoring  (§  56),  we 
have 

(x  +p  +  VqTf)(x  +p  -  VqTf)  =  0  (3) 

This  equation  can  be  satisfied  only  in  two  wajrs,  —  first,  by 
attributing  such  a  value  to  x  as  shall  render  the  first  factor 
equal  to  0 ;  or,  second,  by  attributing  such  a  value  to  x  as 
shall  render  the  second  factor  equal  to  0. 

Placing  the  second  factor  equal  to  0,  we  have 


x+p—  Vq+p^  =  0  ;  and  x'  ^  —p  +  Vq  +p"       (4) 


230  ELEMENTARY    ALGEBRA. 

Placing  the  first  factor  equal  to  0,  we  have 

X+P  +  •\/'q+f  =  0  ;  and  re"  =  —p  —  ^q~+f      (P) 

Since  every  supposition  that  will  satisfy  Equation  (3)  will 
also  satisfy  Equation  (1),  from  which  it  was  derived,  it  follows 
that  x^  and  rr"  are  roots  of  Equation  (1)  ;  also  that 

Every  equation  of  the  secoiid  degree  has  two  roots,  and 
only  two. 

Note.  —  The  two  roots  denoted  by  x^  and  x^^  are  the  same  as  found  in 
2  158. 

Second  Property. 

162.  "We  have  seen  (§  161)  that  every  equation  of  the  sec- 
ond degree  may  be  placed  under  the  form 

{X+P  +  -Vq+p'Xx  +p  -  -Vq+p')  =  0. 

By  examining  this  equation,  we  see  that  the  first  factor  may 
be  obtained  by  subtracting  the  second  root  from  the  unknown 
quantity  x;  and  the  second  factor,  by  subtracting  the  ^irs^ 
root  from  the  unknown  quantity  x.     Hence 

Every  equation  of  the  second  degree  may  he  resolved  into 
two  binomial  factors  of  the  first  degree;  the  first  tei^ms  in  both 
factors  being  the  unkiioivn  quantity,  and  the  second  terms,  the 
roots  of  the  equation  taken  with  contrary  signs. 

Third  Property. 

163.  If  we  add  Equations  (4)  and  (5),  §  161,  we  have 


x'  =- 
x''  =  - 

-  p 

-  p 

X' 

'+«"  =  - 

-2p, 

that  is, 

EQUATIONS  OF  THE  SECOND  DEGREE.         231 

In  every  reduced  equation  of  the  second  degree  the  sum  of 
the  two  roots  is  equal  to  the  coefficient  of  the  second  term  taken 
with  a  contrary  sign. 

Fourth  Prope^'ty. 

164.  If  we  multiply  Equations  (4)  and  (5),  §  161,  member 
by  member,  we  have 


x^  X  a;"  =  (—p  +  ^q  +p'){-p  —  -\/q  +p'') 
^p^  —  (g  -^p^)  ^  ~  ^  /  "*^^^t  is. 
In  every  reduced  equation  of  the  second  degree  the  product 
of  the  two  roots  is  equal  to  the  second  member  taken  with  a 
contrary  sign. 

Formation  of  Equations  of  the  Second  Degree. 

165.  By  taking  the  converse  of  the  second  property  (§  162), 
we  can  form  equations  which  shall  have  given  roots ;  that  is, 
if  they  are  known,  we  can  find  the  corresponding  equations 
by  the  following  rule  :  — 

Subtract  each  root  from  the  unknown  quantity. 
Multiply  the  results  together,  and  place  their  product  equal 
toO. 

Exercises. 

Note.  —  Let  the  pupil  prove,  in  every  case,  that  the  roots  will  satisfy 
the  third  and  fourth  properties. 

1.  If  the  roots  of  an  equation  are  4  and  —5,  what  is  the 

equation  ?  Ans.  rr^  +  rr  =  20. 

2.  What  is  the  equation  when  the  roots  are  1  and  —  3  ? 

Ans.  x^  +  2x=-B. 

3.  What  is  the  equation  when  the  roots  are  9  and  —  10? 

Ans.  x''  +  x-=90. 


282  ELEMENTARY    ALGEBRA. 

4.  What  is  the  equation  whose  roots  are  6  and  —  10? 

Ans.  x^  +  4:x=  60. 

5.  What  is  the  equation  whose  roots  are  4  and  --  3  ? 

Ans.  x^  —  x=  12. 

6.  What  is  the  equation  whose  roots  are  10  and -? 

Ans.  x^  —  9j\a:=  1. 

7.  What  is  the  equation  whose  roots  are  8  and  —  2? 

Ans.  x^  —  6  X  =  16, 

8.  What  is  the  equation  whose  roots  are  16  and  —  5  ? 

Ans.  a:'  — ll:r-=80. 

9.  What  is  the  equation  whose  roots  are  —  4  and  —  5? 

Ans.  .x^  +  9x  =  —  20. 

10.  What  is  the  equation  whose  roots  are  —  6  and  —  7  ? 

Ans.  a:"'  +  13a:  =  — 42. 

3 

11.  What  is  the  equation  whose  roots  are  —  -  and  —  2  ? 

Ans.  x^-{-2^x  =  —  -' 
A 

12.  What  is  the  equation  whose  roots  are  —  2  and  —  3  ? 

Ans.  x^  -\-f>x=^  —  6, 

13.  What  is  the  equation  whose  roots  are  4  and  3  ? 

Ans,  x''-7x  =  —12. 

14.  What  is  the  equation  whose  roots  are  12  and  2? 

Ans.  x^  ~14:X  =  —  24. 

15.  What  is  the  equation  whose  roots  are  18  and  2? 

Ans.  o;'^  — 20 a:  =  —  36. 

16.  What  is  the  equation  whose  roots  are  14  and  3  ? 

Ans.  x''  —  17x-=  -42. 


EQUATIONS    OF   THE    SECOND   DEGREE.  2oP> 

4  9 

17.  What  is  the  equation  whose  roots  are  -  and ? 

^  9  4 

Ans.  2^  A 0;=  1. 

36 
o 

18.  What  is  the  equation  whose  roots  are  5  and ? 

Ans.  Qc^ x  =  — 

3  3 

19.  What  is  the  equation  whose  roots  are  a  and  h  ? 

Ans.  x^  —  {a'\-h)x  =  —  ah. 

20.  What  is  the  equation  whose  roots  are  c  and  —  df 

Ans.  x"^  ~  {c  —  d)x=  cd. 


Trinomial  Equations  of  the  Second  Degree. 

165  a.  A  trinomial  equation  of  the  second  degree  contains 
three  kinds  of  terms  :  — 

First  Term.  —  A  term  involving  the  unknown  quantity  to 
the  second  degree. 

Second  Term.  —  A  term  involving  the  unknown  quantity 
to  the  first  degree. 

Third  Term.  —  A  known  term. 

Thus,  rr^-4:r-12-0 

is  a  trinomial  equation  of  the  second  degree. 

165  b.   What  are  the  factors  of  the  trinomial  equation 

a:'^- 4a: -12  =  0? 

A  trinomial  equation  of  the  second  degree  may  always  be 
reduced  to  one  of  the  four  forms  (§  160)  by  simply  transposing 
the  known  term  to  the  second  member,  and  then  solving  the 
equation.     Thus,  from  the  above  equation  we  have 

x'-^x^\2. 


234  ELEMENTARY    ALGEBRA. 

Resolving  the  equation,  we  find  the  two  roots  to  be  -f-  6 
and  —  2  :  therefore  the  factors  are  x  —  6  and  a:  +  2  (§  162). 

Since  the  sum  of  the  two  roots  is  equal  to  the  coefficient  of 
the  second  term  taken  with  a  contrary  sign  (§  163),  and  the 
product  of  the  two  roots  is  equal  to  the  known  term  in  the 
second  member  taken  with  a  contrary  sign,  or  to  the  third 
term  of  the  trinomial  taken  with  the  sarn.e  sign,  it  follows  that 
any  trinomial  may  be  factored  by  inspection  when  two  num- 
bers can  be  discovered  whose  algebraic  sum  is  equal  to  the 
coefficient  of  the  second  term,  and  whose  product  is  equal  to 
the  third  term. 

Exercises. 

1.  What  are  the  factors  of  the  trinomial  a;^  —  9 5:  —  36  ? 

It  is  seen  by  inspection  that  +  12  and  —  3  will  fulfill  the  conditions 
of  roots  :  for  12  —  3  =  9,  that  is,  the  coefficient  of  the  second  term  with 
a  contrary  sign  ;  and  12  x  —  3  =  -  36,  the  third  term  of  the  trinomial. 
Hence  the  factors  are  a:  —  12  and  a;  +  3. 

2.  What  are  the  factors  of  :r'  -  7.r  —  30  =-  0? 

Ans.  a:  —  10  and  a:  +  3. 

3.  What  are  the  factors  of  :r'  +  15a;  +  36  =  0  ? 

Ans.  x-\-Yl  and  x-\-Z. 

4.  What  are  the  factors  of  .r'  —  12:r  —  28  --  0  ^ 

Ans.  a:  —  14  and  x  -\- 2. 

6.    What  are  the  factors  of  o:^  —  7x  —  8  =  0 ? 

Ans.  X  —  8  and  x  +  I. 

In  the  trinomial  equation  of  the  form  a;-"  +  2joa:"  ~  q,  the  ' 
exponent  of  x  in  the  first  term  is  double  the  exponent  of  x  in 
the  second  term. 

x'~4:x'  =  B2,  and  x*  +  4a;' =  117, 

are  both  equations  of  this  form,  and  may  be  solved  by  the 


EQUATIONS  OF  THE  SECOND  DEGREE.         235 

rules  already  given  for  the  solution  of  equations  of  the  second 
degree. 

In  the  equation       x'^"'  +  2px''  =  q 

we  see  that  the  first  member  will  become  a  perfect  square  by 
adding  to  it  the  square  of  half  the  coefficient  of  x"".     Thus, 

x'''  +  2px''+p^  =  q+p\ 

in  which  the  first  member  is  a  perfect  square.     Then,  extract- 
ing the  square  root  of  both  members,  we  have 


'+p  =  ±:^q+p\ 


Hence  x'^  =  —p  ±  Vq  +  p'^. 

Bv  takincr  the  nth  root  of  both  members, 


x'  -=  ^-p-\--\/q+p\ 


md  x''  ^  \  —  p  —  V—  p  +  p^. 

Exercises. 

1.  What  are  the  values  of  x  in  the  equation 

:^^6-j- 6x^-112? 
Completing  the  square, 

a;6  +  6a;»  +  9  =  112  +  9  =  121. 
Extracting  the  square  root  of  both  members, 

0-3  +  3  =  ±  Vm  =  ±ii. 
Hence  x'  =  ^-3  +  11,  and  x^^  =  \/-3-ll. 

Hence  x^  =  v^8  =  2,        and  x^'  =  v/^H  =  -  v/R 

2.  What  are  the  values  of  x  in  the  equation  x^  —  '^x^  =  9  ? 

Completing  the  square,  we  have 

a;4_8a;2  +  16  =  9  +  16  =  25. 


236  ELEMENTARY    ALGEBRA. 

Extracting  the  square  root  of  both  members, 

a;2  -  4  =  ±  V25  =  ±  5. 
Hence  x^  =  ±  V4  +  5,  and  x^^  =  ±  V4  —  5. 

Hence  a;^  =  -f  3  and  —  3  ;  and  x^^  =  +  V-  i  and  —  V—  1. 

3.    What  are  the  values  of  x  in  the  equation  x^  +  20 o:^  =  69  ? 
Completing  the  square, 

ips  +  20a^  +  100  =  69  4- 100  =  169. 
Extracting  the  square  root  of  both  members, 

(T^  +  10  =  ±  \/l69  =  ±  13. 


t  Hence  x'  =  v/-  10  +  13,  and  x^^  =  V-  10  -  13. 

Hence  x^  =  v^3,  and  o;^^  =  v^^^^. 

4.  What  are  the  values  of  x  in  the  equation  a;*  —  2a:^  —  3  ? 

^Tis.  a;'  =  ±  V3,  and  :r"  =  ±  V—  1. 

5.  What  are  the  values  of  x  in  the  equation  a;^  +  So:^  =  9  ? 

Arts,  a;'  —  1,  and  a;"  =  V— 9. 


6.  Given  a;  ±  V9a:  +  4  =  12,  to  find  a;. 
Transposing  x  to  the  second  member,  and  squaring, 

9a;  +  4  =  rc^-24x  +  144. 
.-.  a;2-33a;  =  -140, 
and  x^  =  28,  and  x^^  =  5. 

7.  4:r±4VJ+2  =  7.  Ans.  x' =  ^,  a:"  =  -. 

4 


8.   x±:-V^x+lO  =  8.  Ans.  x'  =  18,  x"  =  3. 


EQUATIONS    OF    THE   SECOND    DEGREE.  237 

Numerical  Values  of  the  Roots. 

166.  We  have  seen  (§  160)  that,  by  attributing  all  possible 
signs  to  2^  and  <i,  we  have  the  four  following  forms  :  — 

x"  +  2joa;  =  q  (1) 

x^  -2px  =  q  (2) 

x"  +  2px  -=  —  q  (3) 

x^  —  2px  =  —  q  (4) 

Mrst  Form. 

167.  Since  q  is  positive,  we  know,  from  the  Fourth  Prop- 
erty, that  the  product  of  the  roots  must  be  negative :  hence 
the  roots  have  contrary  signs.  Since  the  coefficient  2p  is  posi- 
tive, we  know,  from  the  Third  Property,  that  the  algebraic 
sum  of  the  roots  is  negative  :  hence  the  negative  root  is  numer- 
ically  the  greater. 

Second  Form, 

168.  Since  q  is  positive,  the  product  of  the  roots  must 
be  negative :  hence  the  roots  have  contrary  signs.  Since  2p 
is  negative,  the  algebraic  sum  of  the  roots  must  be  positive : 
hence  the  positive  root  is  numerically  the  greater. 

Third  Form. 

169.  Since  q  is  negative,  the  product  of  the  roots  is  positive 
(Fourth  Property) :  hence  the  roots  have  the  same  sign.  Since 
2p  is  positive,  the  sum  of  the  roots  must  be  negative :  hence 
both  are  negative. 

Fourth  Form. 

170.  Since  q  is  negative,  the  product  of  the  roots  is  positive : 
hence  the  roots  have  the  same  sign.  Since  2p  is  negative,  the 
sum  of  the  roots  is  positive  :  hence  the  roots  are  both  positive. 


238  ELEMENTARY    ALGEBRA. 

171.  If  we  make  ^^  =  0,  the  first  form  becomes 

x"  +  2px  =  0,  or  x(x  +  2p)  =  0] 

which  shows  that  one  root  is  equal  to  0,  and  the  other  to  —  2p. 
Under  the  same  supposition,  the  second  form  becomes 

x"^  —  2px  —  0,  or  x(x  —  2p)  =  0 ; 

which  shows  that  one  root  is  equal  to  0,  and  the  other  to  2p. 
Both  of  these  results  are  as  they  should  be ;  since,  when  q, 
the  product  of  the  roots,  becomes  0,  one  of  the  factors  must 
be  0,  and  hence  one  root  must  be  0. 

172.  If,  in  the  Third  and  Fourth  Forms,  q  >  p'^,  the  quantity 
under  the  radical  sign  will  become  negative :  hence  its  square 
root  cannot  be  extracted  (§  137).  Under  this  supposition,  the 
values  of  x  are  imaginary.  How  are  these  results  to  be 
interpreted? 

If  a  given  nuraher  he  divided  into  two  parts,  their  product 
ivill  be  the  greatest  possible  when  the  parts  are  equal. 

Denote  the  number  by  2p,  and  the  difference  of  the  parts 
by  d.     Then 

p  +  -  =  the  greater  part  (p.  114), 

p  —  -=  the  less  part, 

d^ 
and  p^  -  —  z=  p^  their  product. 

It  is  plain  that  the  product  P  will  increase  as  d  diminishes, 
and  that  it  will  be  the  greatest  possible  when  c?=  0,  for  then 
there  will  be  no  negative  quantity  to  be  subtracted  from  p"^  in 
the  first  member  of  the  equation.  But  when  d=0,  the  parts 
are  equal :  hence  the  product  of  the  two  parts  is  the  greatest 
when  thei/  are  equal. 


EQUATIONS    OF   THE    SECOND    DEGREE.  239 

In  the  equations 

x^  +  2px  =  —  q,     x^  —  2px  =  —  q, 

2p  is  the  sum  of  the  roots,  and  —  q  their  product  ;  and  hence, 
by  the  principle  just  established,  the  product  q  can  never 
be  greater  than  p\  This  condition  fixes  a  limit  to  the  value 
of  q.  If,  then,  we  make  q  >J9^,  we  pass  this  limit,  and  ex- 
press by  the  equation  a  condition  which  cannot  be  fulfilled ; 
and  this  incompatibility  of  the  conditions  is  made  apparent 
by  the  values  of  x  becoming  imaginary.  Hence  we  conclude 
that 

When  the  values  of  the  unlcnown  quantity  are  imaginary, 
the  conditions  of  the  proposition  are  incompatible  with  each 
other. 

Problems  for  Solution. 

1.  Find  two  numbers  whose  sum  shall  be  12,  and  prod- 
uct 46. 

Let  X  and  y  =  the  numbers. 

By  the  1st  condition,  re  +  y  =  12, 

and  by  the  2d,  xy  =  46. 

The  first  equation  gives  x=12—y. 

Substituting  this  value  for  x  in  the  second,  we  have 

12?/ -3/2  =  46. 
Changing  the  signs  of  the  terms,  we  have 

t/2_12y  =  -46. 
Completing  the  square, 

2/2  -  12?/  +  36  =  -  46  +  36  =  -  10, 
which  gives  i/''  =  6  +  V—  10, 

and  2///  =  6  -  V- 10  ; 

both  of  which  values  are  imaginary,  as  indeed  they  should  be,  since  the 
conditions  are  incompatible. 


240  ELEMENTARY    ALGEBRA. 

2.  The  sum  of  two  numbers  is  8,  and  their  product  20. 
What  are  the  numbers  ? 

Let  X  and  y  =  the  numbers. 

By  the  Ist  condition,  x  +  y  =  S. 

By  the  2d,  xy  =  20. 

The  first  equation  gives  x  =  S  —y. 

Substituting  this  value  of  x  in  the  second,  we  have 

8y-2/^=20. 

Changing  the  signs,  and  completing  the  square,  we  have 

i/2-8y  +  16  =  -4. 
Extracting  the  root, 

y'  =  4  -}-  V^,  and  y^^  =  4:  -  V^. 

These  values  of  y  may  be  put  under  the  forms  (§  142), 

y  =  4  +  2\/^,  and  3/  =  4  -  2>/^. 

3.  What  are  the  values  of  a;  in  the  equation  a:*  +  2a;=  — 10? 

Ans.   J-'— 1  +  3V-1. 


1  .^"  : 


4.    Find  a  number  such  that  twice  its  square  added  to  three 
times  the  number  shall  give  65. 

Let  a;  —  unknown  number. 

Then  the  equation  of  the  problem  is 

2x2  + 3a;  =  65. 


Whence              x  = 

=-!m 

'2       16 

-h 

23 
4* 

..-5 

*!-=■ 

and  x^^  « 

3 
4 

23 
4  "^ 

13 

'  2* 

Both  these  values 

satisfy  the  equation 

of  th( 

3  probh 

im  : 

for 

2  X  (5)'' 

+  3x5- 

2  X  25  +  15  = 

=  65, 

and  2('-23Y  +  3x-I5  =  l^-3?-130.65 

\      2  /  2         2         2         2 


EQUATIONS    OF    THE    SECOND    DEGREE.  241 

Notes.  —  1.  If  we  restrict  the  enunciation  of  the  problem  to  its 
arithmetical  sense,  in  which,  "added"  means  "numerical  increase,"  the 
first  value  of  x  only  will  satisfy  the  conditions  of  the  problem. 

2.  If  we  give  to  "  added  "  its  algebraical  signification  (when  it  may 
mean  subtraction  as  well  as  addition),  the  problem  may  be  thus  stated :  — 

To  find  a  number  such  that  twice  its  square  diminished  by  three 
times  the  number  shall  give  65. 

The  second  value  of  x  will  satisfy  this  enunciation ;  for 

\2)  2        2        2 

3.  The  root  which  results  from  giving  the  plus  sign  to  the  radical  is 
generally  an  answer  to  the  question  in  its  arithmetical  sense.  The 
second  root  generally  satisfies  the  problem  under  a  modified  statement. 

Thus,  in  the  example  it  was  required  to  find  a  number  of  which  twice 
the  square  added  to  three  times  the  number  shall  give  65.  Now,  in  the" 
arithmetical  sense,  "  added  "  means  increased ;  but  in  the  algebraic  sense 
it  implies  diminution  when  the  quantity  added  is  negative.  In  this 
sense,  the  second  root  satisfies  the  enunciation. 

6.  A  certain  person  purchased  a  number  of  yards  of  cloth 
for  240  cents.  If  he  had  purchased  3  yards  less  of  the  same 
cloth  for  the  same  sum,  it  would  have  cost  him  4  cents  more 
per  yard.     How  many  yards  did  he  buy  ? 

Let  X  =  number  'of  yards  purchased. 

Then  —  =  the  price  per  yard. 

X 

If  for  240  cents  he  had  purchased  three  yards  less,  that  is,  ar  —  3 
yards,  the  price   per   yard,  under   this   hypothesis,  would   have   been 

240 

denoted  by  ;-•     But  by  the  conditions  this  last  cost  must  exceed  the 

a:  —  3 
first  by  4  cents.     Therefore  we  have  the  equation 

240       240  _  ^ 


x-3        X 

Whence 

x2-3a;=180, 

and 

.  =  |.V|-^^^o  =  ^f- 

.-.  x'  =-- 15,  and  x''  =  -  12. 

D. 

y.  E.  A.—  16. 

242  ELEMENTARY    ALGEBRA. 

Notes.  —  1.  The  value  x^  =  15  satisfies  the  enunciation  in  its  arith- 
metical sense  ;  for  if  15  yards  cost  240  cents,  240  -j-  15  =  16  cents,  the 
price  of  one  yard ;  and  240  -?- 12  =  20  cents,  the  price  of  1  yard  under 
the  second  supposition. 

2.   The  second  value  of  x  is  an  answer  to  the  following  problem  :  — 

A  certain  person  purchased  a  number  of  yards  of  cloth  for  240  cents. 
If  he  had  paid  the  same  for  three  yards  more,  it  would  have  cost  him  4 
cents  UiB,  per  yard.     How  many  yards  did  he  buy  ? 

This  would  give  the  equation  of  condition 
240  _  _240_  _  ^ 
X       a;  +  3 
or  a;2-3a;=180, 

the  same  equation  as  found  before.     Hence 

A  single  equation  will  often  state  two  or  more  arithmetical  problems. 

This  arises  from  the  fact  that  the  language  of  algebra  is  more  com- 
prehensive than  that  of  arithmetic. 

6.  A  man,  having  bought  a  horse,  sold  it  for  $24.  At  the 
sale  he  lost  as  much  per  cent  on  the  price  of  the  horse  as  the 
horse  cost  him  dollars.     What  did  he  pay  for  the  horse  ? 

Let  X  =  the  number  of  dollars  that  he  paid  for  the  horse.  Then 
a;  —  24  =  the  loss  he  sustained.     But  as  he  lost  x  per  cent  by  the  sale, 

he  must  have  lost  -^  upon  each  dollar,  and  upon  x  dollars  he  lost  a 
100    ^  ^ 

sum  denoted  by  - — .     We  have,  then,  the  equation 
^  100  ^ 

—  =  re  -  24,  whence  x'^  -  100  a;  =  -  2400, 
100 


and  a;  =  50  ±  V2500  -  2400  =  50  ±  10. 

.-.  x^  =  60,  and  x'^  =  40. 

Both  of  these  roots  will  satisfy  the  problem  :  for,  if  the  man  gave 
$60  for  the  horse,  and  sold  him  for  $24,  he  lost  $36.  From  the  enun- 
ciation he  should  have  lost  60  per  cent  of  $60  ;  that  is, 

i0^f60  =  ?^^<^^  =  36. 
100  100 

Therefore  $60  satisfies  the  enunciation. 


EQUATIONS   OF    THE    SECOND    DEGREE.  243 

Had  he  paid  $40  for  the  horse,  he  would  have  lost  by  the  sale  $16. 
From  the  enunciation  he  should  have  lost  40  per  cent  of  $40  ;  that  is, 

100  100 

Therefore  $40  satisfies  the  enunciation. 

7.  The  sum  of  two  numbers  is  11,  and  the  sum  of  their 
squares  is  61.     What  are  the  numbers?  Ans.  5  and  6. 

8.  The  difference  of  two  numbers  is  3,  and  the  sum  of  their 
squares  is  89.     What  are  the  numbers  ?  Ans.  5  and  8. 

9.  A  grazier  bought  as  many  sheep  as  cost  him  £60 ;  and, 
after  reserving  fifteen  out  of  the  number,  he  sold  the  re- 
mainder for  £54,  and  gained  2s.  a  head  on  those  he  sold. 
How  many  did  he  buy  ?  Ans,  75. 

10.  A  merchant  bought  cloth,  for  which  he  paid  £33  15s., 
which  he  sold  again  at  £2  8s.  per  piece,  and  gained  by  the 
bargain  as  much  as  one  piece  cost  him.  How  many  pieces 
did  he  buy  ?  Ans.  15. 

11.  The  difference  of  two  numbers  is  9,  and  their  sum 
multiplied  by  the  greater  is  equal  to  266.  What  are  the 
numbers  ?  ,  Ans.  14  and  5. 

12.  Find  a  number  such  that,  if  you  subtract  it  from  10 
and  multiply  the  remainder  by  the  number  itself,  the  product 
will  be  21.  Ans.  7  or  3. 

13.  A  person  traveled  105  miles.  If  he  had  traveled  2 
miles^an  hour  slower,  he  would  have  been  6  hours  longer 
in  completing  the  same  distance.  How  many  miles  did  he 
travel  per  hour  ?  Ans.  7  miles. 

14.  A  person  purchased  a  number  of  sheep,  for  which 
he  paid  $224.     Had  he  paid  for  each  twice  as  much,  plus 


244  ELEMENTARY    ALGEBRA. 

$2,  the  number  bought  would  have  been  represented  by 
twice  what  was  paid  for  each.  How  many  sheep  were  pur- 
chased? Ans.  32. 

15.  The  difference  of  two  numbers  is  7,  and  their  sum 
multiplied  by  the  greater  is  equal  to  130.  What  are  the 
numbers?  Ans.  10  and  3. 

16.  Divide  100  into  two  such  parts  that  the  sura  of  their 
squares  shall  be  5392.  Ans.  64  and  36. 

17.  Two  square  courts  are  paved  with  stones  a  foot  square. 
The  larger  court  is  12  feet  larger  than  the  smaller  one,  and 
the  number  of  stones  in  both  pavements  is  2120.  How  long 
is  the  smaller  pavement  ?  Ans.  26  feet. 

18.  Two  hundred  and  forty  dollars  are  equally  distributed 
among  a  certain  number  of  persons.  The  same  sum  is  again 
distributed  amongst  a  number  greater  by  4.  In  the  latter 
case  each  receives  ten  dollars  less  than  in  the  former.  How 
many  persons  were  there  in  each  case  ?  Ans.  8  and  12. 

19.  Two  partners,  A  and  B,  gained  $360.  A's  money  was 
in  trade  12  months,  and  he  received  for  principal  and  profit 
$520.  B's  money  was  $600,  and  was  in  trade  16  months. 
How  much  capital  had  A  ?  Ans.  $400. 

Equations  containing  more  than  One  Unknown 
Quantity. 

173.  Two  simultaneous  equations,  each  of  the  second  de- 
gree, and  containing  two  unknown  quantities,  will,  when 
combined,  generally  give  rise  to  an  equation  of  the  fourth 
degree.  Hence  only  particular  cases  of  such  equations  can 
be  solved  by  the  methods  already  given. 


EQUATIONS   OF   THE   SECOND    DEGREE.  245 

Case  I. 

Two  simultaneous  equations,  involving  two  unknown  quanti- 
ties, may  readily  he  solved  when  o?ie  is  of  the  first  and  the  other 
of  the  second  degree. 

(1)    Given    ['^^  +  y^Z^^],i^i.n^:c^n^y. 

^^•12,.'=  6,    y'=8. 

Transposing  y  in  the  first  equation,  we  have 

(B  =  14  —  ?/. 

Squaring  both  members,  x^  =  196  —  28y  +  y"^. 

Substituting  this  value  for  x^  in  the  second  equation,  we  have 

196 -282/ +  2/' +  3/' =  100, 

from  which  we  have  y"^  -l^y  = —  ^Bi. 

Completing  the  square, 

2/2 -My +  49  =  1. 

Extracting  the  square  root, 

y~1  =  ±  VI  =  +  1  and  -  1. 

Hence  2/^=7  +  1  =  8, 

and  2/^/=  7  — 1  =  6. 

If  we  take  the  greater  value,  we  find  a:  =  6  ;  and  if  we  take  the  lesser, 
we  find  a;  =  8. 

Verification.     For  the  greater  value,  y  =^  8,  the  equation 
a;  -f  2/  =  14 
gives  6  +  8  =  14, 

and  a;'  +  2/2  =  100 

gives  36  +  64  =  100. 


246  ELEMENTARY    ALGEBRA. 

For  the  value  3/  =  6,  the  equation 

X  +  y  =  14: 
gives  8  +  6  =  14, 

and  x'^  +  y^==100 

gives  64  +  36  =  100. 

Hence  both  sets  of  values  satisfy  the  given  equation. 

(2)    Given    J^,~^.^  =  J|,tofind.randy. 

Transposing  y  in  the  first  equation,  we  have 

x  =  3  +y. 
Squaring  both  members,  x"^  =  d  -{■  6y  +  y^. 

Substituting  this  value  for  x^  in  the  second  equation,  we  have 
9  +  62/  +  3/'»-7/2  =  45. 

Whence  6y  =  36,  and  y  =  6. 

Substituting  this  value  of  y  in  the  first  equation,  we  have 

a;  -  6  =  3, 

and  consequently  a;''  =  3  -|-  6  =  9. 

Verification. 

X  —  y  =    3  gives     9  —   6  =    3, 

and  x^-y^  =  45  gives  81  -  36  =  45. 

Exercises. 
Solve  the  following  simultaneous  equations  :  — 
1. 


2. 
3. 

4. 


fcr  +y  =12) 
U'  -  y»  =  24  j 

(.  y'  =  5, 

U''  +  y  =  117j 

-^^   fa;' =  9,  a;"  =  -6. 

f^+y  =  9           ) 

\x'-2xy  +  y'  =  l^ 

ty=4,  y"=4. 

U;'  +  2a-y  +  y'  =  225J 

,        fa:' =  10,  x"=-    5. 

^'^My^  5,y'=-io. 

EQUATIONS   OF    THE    SECOND    DEGREE.  247 

Case  II. 

174,  Two  simultaneous  equations  of  the  second  degree,  which 
are  homogeneous  with  respect  to  the  unJcnoion  quantity,  may 
readily  be  solved. 

Gxven  |^^  +  3^y  =  22  (1) 


to  find  X  and  y. 


Assume  x  =^ty,  t  being  any  auxiliary  unknown  quantity. 
Substituting  this  value  of  x  in  Equations  (1)  and  (2),  we  have 

ty  +  3tf  =  22  ,.  y»  =  -^  (3) 

40 


Hence 


40 


f'  +  St     ^2  ^  3 i  +  2 
Hence  22  i^  +  66  i  +  44  -  40  ^2  +  120 1. 

Reducing,  f  -{■  3t  =  — 

Whence,  t'  =  -,  and  t''  =  -  li. 

3  3 

Substituting  either  of  these  values  in  Equations  (3)  or  (4),  we  find 

2/^  =  +  3,  and  i/^  =  -  3. 

Substituting  the  plus  value  of  y  in  Equation  (1),  we  have 

.T2-}-9a;  =  22, 
from  which  we  find 

a;^  =  +  2,  anda^^^  =  -ll. 

If  w^  take  the  negative  value,  y^^  =  -  3,  we  have  from  Equation  (1) 

a;2-9a;  =  22, 

from  which  we  find  o;^  =  +  11.  and  .r^^  =  —  2, 


248  jELEMENTARY    ALGEBRA. 

Verification.  For  the  values  y^  =  +  3,  and  x^  =  -f  2,  the  given 
equation 

gives  22 +  3x2x3  =  4  + 18  =  22; 

and  for  the  second  value,  x^^  =  —  11,  th»  same  equation, 

a;2  +  3a;y  =  22, 

gives  (-  11)2  +  3  X  -  11  X  3  =  121  -  99  =  22. 

If,  now,  we  take  the  second  value  of  y  (that  is,  y^^  =  —  3)  and  the 
corresponding  values  of  x  (viz.,  x^  =  +  11,  and  x^^  =  —  2)  for  x^  =  -\- 11, 
the  given  equation, 

a;2  +  3a?2/  =  22, 

gives  IP  +  3  X  11  X  -  3  =  121  ~  99  =  22  ; 

and  for  x^^  =  —  2,  the  same  equation, 

a;2  +  3a;y  =  22, 

gives  (-  2)2  +  3  X  -  2  X  -  3  =  4  +  18  =  22. 

The  verifications  could  be  made  in  the  same  way  by  employing 
Equation  (2). 

Note.  —  In  equations  similar  to  the  above  we  generally  find  but  a 
single  pair  of  values,  corresponding  to  the  values  in  this  equation  of 
y'  =  +  3,  and  x^  =  -\-  2.  The  complete  solution  would  give  four  pairs 
of  values. 

Exercises. 

Find  the  values  of  x  and  y  in  the  following  equations :  — 

K.y'  -~xy=      b)  iy  =  b. 

\f  +ar'=    85j  (y  =  7. 

{    y"-    xy=-^)  ly=   9. 

^      \bxy-Sf       =32)  ,        {x=l. 

U'+y'  +  3a;2/=71  j  '  ly=l. 


EQUATIONS    OF   THE   SECOND    DEGREE.  249 

Case  III. 

175.  Many  other  equations  of  the  second  degree  may  be  so 
transformed  as  to  be  brought  under  the  rules  of  solution 
already  given.  The  seven  formulas  following  will  aid  in 
such  transformation. 

Formula  1.  —  When  the  sum  and  the  diflference  are  known. 
x  +  y  =  s. 
X  —  y  =^  d. 

Then  [p.  125,  (3)],      x^'^=^s  +  U, 
and  y^---  =  -s--d, 

Pormula  2.  —  When  the  sum  and  the  product  are  known. 

(1) 

(2) 


Squaring  (1), 
Multiplying  (2)  by  4, 

x  +  y^s 
xy^p 

x^  +  2xy  +  y''-=s'' 
4txy         =4ip 

Subtracting, 

x^-2xy  +  y^  =  s^-^p 

Extracting  root, 
But 

x  —  y  =  ±  Vs'  —  4^. 
x  +  y  =  s. 

Hence 

1 

and 

l/  =  l^l^^'-^P- 

250  ELEMENTARY    ALGEBRA. 

Formula  3.  —  When  the   difference   and   the   product  are 
known. 


Squaring  (1), 
Multiplying  (2)  by  4, 

X  —  y  ^=  d 

x^-2xy  +  f  =  d' 
^xy          =  4jo 

(1) 

(2) 

Adding, 

x:^  +  2xy  +  f  =  d'  +  4:p 

x  +  y  =  ±L  -y/d'^  +  ^P 
x-y^-d 

1         1 

X     =      ^d±^y/d'  +  ^p 

Formnla  4.  —  When  the  sum  of  the  squares  and  the  product 
are  known. 

x'  +  f^s  (1) 

xy  --=p  (2) 

.-.  2xy^2p  (3) 

Adding  (1)  and  (3), 


Hence  x  +  y  —  i  Vs  +  2^  (4) 

Subtracting  (3)  from  (1), 

x^  —  2xy  +  3/2  ~  s  —  2p, 


Hence  x  —  y=^±-\/s  —  2p  (5) 

Combining  (4)  and  (5),  x  =  -Vs  +  2p  +  - Vs  —  2^, 

^  -J 

and  y  =  -  Vs  +  2jo  —  -  Vs  —  2p. 


EQUATIONS    OF    THE    SECOND    DEGREE.  251 

rormula  5. — When  the  sum,  and  the  sum  of  the  squares, 
are  known. 

x  +  y  =  s  (1) 

x^  +  y'  =  s^  (2) 

Squaring  (1),        x^  +  2xy  +  y'^  =  s^ 

2xy  =  s^  —  s' 

r  —  s' 
xy  =  ~-=p.  (3) 

By  putting  xy=^p,  and  combining  Equations  (1)  and  (3) 
by  Formula  2,  we  find  the  values  of  x  and  y. 


Formula  6. - 

-When 

the 

\  sum,  and  the  sum  of  the  cubes, 

are  known. 

x  +  y  =  S                          (1) 
a.-'  +  y'=152                        (2) 

Cubing  (1), 

x^  +  ^ 

x'^y 

+  3:ry'  +  y»=512 

Subtracting, 

3x^4- 3a:y»  =  360 

Factoring, 

Zxy{x  +  y)^Zm 

From  (1), 

32ry(8)  =  360 
24a;y  =  360 

He~nce 

xy  =  15                        (3) 

Combining  (1)  and  (3),  we  find  a;  =  5,  and  y  =  3. 

Formula  7.  —  When  we  have  an  equation  of  the  form 

{x  +  yY  +  {x  +  y)  =  q. 
Let  us  assume  x-\-y  ^=z. 

Then  the  given  equation  becomes 

'      2J'  +  2  =  5^,  and  z  =  -  -  ±y^q  +  -• 


'+y^-\^\^  +  \ 


252  ELEMENTARY    ALGEBRA. 

r  xz  =  y'     (l)^ 

(1)  Given    }  x  +y  +z  =    7     (2)  >- ,  to  find  x,  y,  and  z. 
Lx'  +  y'  +  z'  =  2l     (3)) 

Transposing  y  in  (2),  we  have 

x  +  z  =  7  -y  (4) 

Squaring  the  members,  we  have 

a;2  +  2x2  +  2^  =  49  -  14y  +  y*. 
If,  now,  we  substitute  for  2xz  its  value  taken  from  (1),  we  have 

a;2  +  2?/2  +  22  =  49  -  142/  +  y'^'     • 
Canceling  y^  in  each  member,  there  results 

x^  +  y^  +  z^^Ad-Uy. 

But  from  (3)  we  see  that  each  member  of  the  last  equation  is  equal 

to  21.    Hence 

49-14y  =  21, 

and  '        142/ =  49 -21  =  28. 

Hence  y  =  —  =  2. 

14 

Substituting  this  value  of  y  in  (1)  gives 

xz  =  4t. 
Substituting  it  in  (4)  gives 

X  -\-  z  =  5,  or  X  =  5  —  z. 

Substituting  this  value  of  x  in  the  previous  equation,  we  obtain 

5  2  —  2^  =  4,  or  2^  —  5  2  =  —  4. 

Completing  the  square,  we  have 

22-52  +  6.25  =  2.25; 

and  2  -  2.5  =  i:  VT25  =  +  1.5  or  -  1.5. 

Hence  2  =-  2.5  +  1.5  =  4,  and  2  =  +  2.5  -  1.5  =  1. 


2a;  +  2y  = 

=  26 

a;+    y  = 

=  13 

Vxy  +  13  = 

a9 

y/xy  = 

=    6 

xy  = 

:36. 

a;2+    xy  +  y^  = 

:133 

3xy 

=  108 

EQUATIONS   OF   THE    SECOND    DEGREE.  253 

{2)GWen\\+^  +  ^r,ll     JJ^l.tofindxandy. 
I  o;'  +      rr^  +  y'^  =  133     (2)  ) 

Dividing  (2)  by  (1),  we  have 

X  —  Vxy  +    y  =    7 

But  X  +  Vxy  +    2/  =  19 

Hence,  by  addition, 
or 

Substituting  in  (1), 
or,  transposing. 

Squaring, 

(2)  is 
and  from  the  last  we  have 

Subtracting,  x"^  —  2xy  +  y'^=    25 

Hence  x  —  y  =  ±5 

But  x-\-y  =  lS 

Hence  x  =  9  and  4,  and  2/  =  4  and  9. 


Problems  for  Solution. 

1.    Find  two  numbers  sucli  that  their  sum  shall  be  15,  and 
the  sum  of  their  squares  113. 

Let  X  and  y  =  the  numbers. 

Then  x  +  y  =  15  (1) 

and  x^  +  y^  =  m  (2) 

From  (1)  we  have  x^  =  225  -  30y  +  y\ 

Substituting  this  value  in  (2), 

225 -302/ +  2/2 +  2/2  =  113. 

Hence  2y^-30y  =  -U2, 

and  2/2 -152/ =  -56. 

Hence  y^  =  8,  and  y^^  =  7. 

The  first  value  of  y  being  substituted  in  (1)  gives  x^  =7  ]  and  the 
second,  x/^  =  8.     Hence  the  numbers  are  7  and  8. 


254  ELEMENTARY    ALGEBRA. 

2.  Find  two  numbers  such  that  their  product  added  to 
their  sum  shall  be  17,  and  their  sum  taken  from  the  sum  of 
their  squares  shall  leave  22. 

Let  X  and  y  =  the  numbers. 

Then,  from  the  conditions, 

{x  +  y)  +  xy^l7  (1) 

x^-\-y^-{x-\-y)  =  22  (2) 

Multiplying  (1)  by  2,  we  have 

2a;?/  +  2(a;  +  3/)  =  34  (3) 

Adding  (2)  and  (3),  we  have 

x^  +  2xy  +  2/2  +  {x  +  y)  =  56. 

Hence  (a;  +  7/f  +{x  -\-y)  =  56  (4) 

Regarding  (x  +  y)  as  a  single  unknown  quantity, 


a;  +  y  =  -U-^56  +  i  =  7. 


2      >  4 

Substituting  this  value  in  (1),  we  have 

7  +  iry  =  17,  and  2/ =  5. 
Hence  the  numbers  are  2  and  5, 

3.  What  two  numbers  are  those  whose  sum  is  8,  and  sum 
of  their  squares  34  ?  Ans.  5  and  3. 

4.  It  is  required  to  find  two  such  numbers  that  the  first 
shall  be  to  the  second  as  the  second  is  to  16,  and  the  sum  of 
whose  squares  shall  be  225.  Ans.  9  and  12. 

6.  What  two  numbers  are  those  which  are  to  each  other 
as  3  to  5,  and  whose  squares  added  together  make  1666  ? 

Ans.  21  and  35. 

6.  There  are  two  numbers  whose  difference  is  7,  and  half 

their  product  plus  30  is  equal  to  the  square  of  the  less  num- 
ber.    What  are  the  numbers?  Ans.  12  and  19. 

7.  What  two  numbers  are  those  whose  sum  is  5,  and  the 
sum  of  their  cubes  35  ?  Ans.  2  and  3. 


I 


EQUATIONS    OF    THE   SECOND   DEGREE.  255 

8.  What  two  numbers  are  those  whose  sum  is  to  the  greater 
as  11  to  7,  and  the  difference  of  whose  squares  is  132? 

Ans.  14  and  8. 

9.  Divide  the  number  100  into  two  such  parts  that  the 
product  may  be  to  the  sum  of  their  squares  as  6  to  13. 

Ans.  40  and  60. 

10.  Two  persons,  A  and  B,  departed  from  different  places 
at  the  same  time,  and  traveled  towards  each  other.  On  meet- 
ing, it  appeared  that  A  had  traveled  18  miles  more  than  B, 
and  that  A  could  have  gone  B's  journey  in  15f  days,  but  B 
•would  have  been  28  days  in  performing  A's  journey.  How 
far  did  each  travel?  Ans.  A,  72  miles  ;  B,  54  miles. 

11.  There  are  two  numbers  whose  difference  is  15,  and 
half  their  product  is  equal  to  the  cube  of  the  lesser  number. 
What  are  those  numbers  ?  Ans.  3  and  18. 

12.  What  two  numbers  are  those  whose  sum  multiplied  by 
the  greater  is  equal  to  77,  and  whose  difference  multiplied  by 
the  less  is  equal  to  12  ?       ^^^^_  ^  ^^^  ^^  ^^  3  ^g  ^„a  ^  V2. 

13.  Divide  100  into  two  such  parts  that  the  sum  of  their 
iquare  roots  may  be  14.  Ans.  64  and  36. 


14.    It  is  required  to  divide  the  number  24  into  two  such 

parts  that  their  product  may  be  equal  to  35  times  their  dif- 

.ference.  Ans.  10  and  14. 

I 

^      15.    The  sum  of  two  numbers  is  8,  and  the  sum  of  their 
cubes  is  152.     What  are  the  numbers?  .4ns.  3  and  5. 

16.    Two  merchants  each  sold  the  same  kind  of  stuff.     The 
second  sold  3  yards  more  of  it  than  the  first,  and  together 


256  ELEMENTARY    ALGEBRA. 

they  receive  $35.  The  first  said  to  the  second,  "  I  would  have 
received  $24  for  your  stuff."  The  other  replied,  '*  And  I 
should  have  received  $12^  for  yours."  How  many  yards  did 
each  of  them  sell  ? 

Ans.  1st  merchant,  15  or  5  ;    2d  merchant,  18  or  8. 

17.  A  widow  possessed  $  13,000,  which  she  divided  into  two 
parts,  and  placed  them  at  interest  in  such  a  manner  that  the 
incomes  from  them  were  equal.  If  she  had  put  out  the  first 
portion  at  the  same  rate  as  the  second,  she  would  have  drawn 
for  this  part  $360  interest ;  and  if  she  had  placed  the  second 
out  at  the  same  rate  as  the  first,  she  would  have  drawn  for  it 
$490  interest.     What  were  the  two  rates  of  interest  ? 

Ans.  7  and  6  per  cent. 

18.  Find  three  numbers  such  that  the  difference  between 
the  third  and  second  shall  exceed  the  difference  between  the 
second  and  first  by  6,  that  the  sum  of  the  numbers  shall  be 
33,  and  the  sum  of  their  squares  467.  Ans.  5,  9,  and  19. 

19.  What  number  is  that  which  being  divided  by  the 
product  of  its  two  digits  the  quotient  will  be  3,  and  if  18  be 
added  to  it  the  resulting  number  will  be  expressed  by  the 
digits  inverted  ?  Ans.  24. 

20.  What  two  numbers  are  those  which  are  to  each  other 
as  m  to  n,  and  the  sum  of  whose  squares  is  b  f 

my/b  n^b 


Ans. 


21.    What  two  numbers  are  those  which  are  to  each  other 
as  m  to  n,  and  the  difference  of  whose  squares  is  b  f 

m-\/b  n^b 


Ans. 


-y/m^  —  n^     V: 


m' 


22.    Required  to  find  three  numbers  such  that  the  product 
of  the  first  and  second  shall  be  equal  to  2,  the  product  of  the 


EQUATIONS    OF    THE   SECOND    DEGREE.  257 

first  and  third  equal  to  4,  and  the  sum  of  the  squares  of  the 
second  and  third  equal  to  20.  Ans.  1,  2,  and  4. 

23.  It  is  required  to  find  three  numbers  whose  sum  shall 
be  38,  the  sum  of  their  squares  634,  and  the  difference 
between  the  second  and  first  greater  by  7  than  the  difference 
between  the  third  and  second.  Ans.  3,  15,  and  20. 

24.  Required  to  find  three  numbers  such  that  the  product 
of  the  first  and  second  shall  be  equal  to  a,  the  product  of  the 
first  and  third  equal  to  b,  and  the  sum  of  the  squares  of  the 
second  and  third  equal  to  c. 


Ans.  yf±^^    ayj-^,    by]- 


a'  +  b''      Ma'  +  b' 

25.  What  two  numbers  are  those  whose  sum  multiplied  by 
the  greater  gives  144,  and  whose  difference  multiplied  by  the 
less  gives  14  ?  Ans,  9  and  7. 

D.  N.  E.  A,  — 17. 


CHAPTER  IX. 

PROPORTIONS    AND    PROGRESSIONS. 

176.  Two  quantities  of  the  same  kind  may  be  compared, 
the  one  with  the  other,  in  two  ways,  —  first,  by  considering 
how  much  one  is  greater  or  less  than  the  other,  which  is  shown 
by  their  difference ;  and,  second,  by  considering  how  many 
times  one  is  greater  or  less  than  the  other,  which  is  shown  by 
their  quotient. 

Thus,  in  comparing  the  numbers  3  and  12  together  with 
respect  to  their  difference,  we  find  that  12  exceeds  3  by  9  ; 
and  in  comparing  them  together  with  respect  to  their  quo- 
tient, we  find  that  12  contains  3  four  times,  or  that  12  is  four 
times  as  great  as  3. 

The  first  of  these  methods  of  comparison  is  called  arithmetical 
proportion;  and  the  second,  geometrical  proportion.     Hence 

Arithmetical  proportion  considers  the  relation  of  quantities 
with  respect  to  their  difference ;  and  geometrical  proportion,  the 
relation  of  quantities  ivith  respect  to  their  quotient. 

Arithmetical  Proportion  and  Progression. 

177.  If  we  have  four  numbers,  2,  4,  8,  and  10,  of  which 
the  difference  between  the  first  and  second  is  equal  to  the 
difference  between  the  third  and  fourth,  these  numbers  are 
said  to  be  in  arithmetical  proportion.  The  first  term,  2,  is 
called  an  antecedent ;  and  the  second  term,  4,  with  w^hich  it 
is  compared,  a  consequent,  The  number  8  is  also  called  an 
antecedent;  and  the  number  10,  with  which  it  is  compared, 
a  consequent. 

5o8 


PROPORTIONS    AND    PROGRESSIONS.  259 

When  the  difference  between  the  first  and  second  is  equal 
to  the  difference  between  the  third  and  fourth,  the  four  num- 
bers are  said  to  be  in  proportion.     Thus,  the  numbers 

2,     4,     8,     10, 

are  in  arithmetical  proportion. 

178.  When  the  difference  between  the  first  antecedent  and 
consequent  is  the  same  as  between  any  two  consecutive  terms 
of  the  proportion,  the  proportion  is  called  an  arithmetical  pro- 
gression. Hence  a  progression  by  differences,  or  an  arithmetical 
progression,  is  a  series  in  which  the  successive  terms  are  con- 
tinually increased  or  decreased  by  a  constant  number,  which 
is  called  the  common  difference  of  the  progression. 

A  series  is  a  succession  of  terms,  each  of  which  is  derived 
from  one  or  more  of  the  preceding  ones  by  2^,  fixed  law,  called 
the  law  of  the  series. 


In  the  two  series, 

1,       4,       7, 

10, 

13, 

16, 

19, 

22, 

25, 

60,     56,     52, 

48, 

44, 

40, 

36, 

32, 

28, 

the  first  is  called  an  increasing  progression,  of  which  the  com- 
mon difference  is  3  ;  and  the  second,  a  decreasing  progression, 
of  which  the  common  difference  is  4. 

In  general,  let  a,  6,  c,  d,  e,f,  denote  the  terms  of  a  pro- 
gression by  differences.     It  has  been  agreed  to  write  them 

thus  :  — 

a  .b  .c  .d  .e  .f  .g  .h  .i  .  h 

This  series  is  read,  "  a  is  to  ^,  as  5  is  to  c,  as  c  is  to  d,  as  d 
is  to  e,"  etc.  This  is  a  series  of  continued  equi- differences,  in 
which  each  term  is  at  the  same  time  an  antecedent  and  a 
consequent,  with  the  exception  of  the  first  term,  which  is  only 
an  antecedent,  and  the  last,  which  is  only  a  consequent. 


260  ELEMENTARY   ALGEBRA. 

179.    Let  d  denote  the  common  difference  of  the  progression, 
a  .h  .c  .e  .f  .g  .  h^  etc., 
which  we  will  consider  increasing. 

From  the  definition  of  the  progression,  it  evidently  follows 
that 

h  =  a-\-  d,    c  =  b  -}-  d  =  a-{-2d,    e  =  c-{-d  =  a-i-Sd; 

and,  in  general,  any  term  of  the  series  is  equal  to  the  first 
term,  plus  as  many  times  the  common  difference  as  there  are 
preceding  terms. 

Thus,  let  I  be  any  term,  and  n  the  number  which  marks 
the  place  of  it.     The  expression  for  this  general  term  is 

I  =  a  +  (n  —  1)  d. 

Hence,  for  finding  the  last  term,  we  have  the  following 
rule  :  — 

Multiply  the  coynmon  difference  by  the  number  of  terms 
less  one. 

To  the  product  add  the  first  term.  The  sum  will  be  the 
last  term. 

The  formula  1=  a-\-{7i—V)d 

serves  to  find  any  term  whatever,  without  determining  all 
those  which  precede  it. 

Exercises. 

1.  If  we  make  n  —  \,  we  have  l  =  a;  that  is,  the  series 
will  have  but  one  term. 

2.  If  we  make  n  =  2,  we  have  l—a^d;  that  is,  the 
series  will  have  two  terms,  and  the  second  term  is  equal  to 
the  first,  plus  the  common  difference. 


PROPORTIONS    AND    PROGRESSIONS.  261 

3.  U  a  =  3,  and  d—2,  what  is  the  3d  term?  Arts.  7. 

4.  If  a  =^  5,  and  d  =  4:,  what  is  the  6th  term  ?  Ans.  25. 

5.  If  a  =  7,  and  d  —  5,  what  is  the  9th  term?  Ans.  47. 

6.  If  a  =  8,  and  d—  5,  what  is  the  10th  term?  Arts.  53. 

7.  If  a  =  20,  and  d  =  4:,  what  is  the  12th  term  ?  Ans.  64. 

8.  If  a  =-  40,  and  d==20,  what  is  the  50th  term? 

Ans.  1020. 

9.  If  a  =  45,  and  d  =  30,  what  is  the  40th  term  ? 

Ans.  1215. 

10.  If  a  =-  30,  and  d  -=  20,  what  is  the  60th  term  ? 

Ans.  1210. 

11.  If  a  --  50,  and  c^  =  10,  what  is  the  100th  term  ? 

Ans.  1040. 

12.  To  find  the  50th  term  of  the  progression 

1.4.7.10.13.16.19 , 

we  have  ^  =  1  +  49  X  3  =  148. 

13.  To  find  the  60th  term  of  the  progression 

1.5.9.13.17.21.25 , 

we  have  Z  =  1  +  59  X  4  =  237. 

180.   If  the  progression  were  a  decreasing  one,  we  should 

have  1  =  a  —  (n—l)d. 

Hence,  to  find  the  last  term  of  a  decreasing  progression, 
we  have  the  following  rule  :  — 

Multiple/  the  common  diffeo^ence  hy  the  number  of  terms 
less  one. 

Subtract  the  product  from  the  first  term.  The  remainder 
ivill  he  the  last  term. 


262  ELEMENTARY    ALGEBRA. 


Exercises. 

1.  The  first  term  of  a  decreasing  progression  is  60,  the 
number  of  terms  20,  and  the  common  difference  3.  What 
is  the  last  term  ? 

l^a-{n-l)d  gives  ^  =  60 -(20  -  J)3  =  60- 57  =  3. 

2.  The  first  term  is  90,  the  common  difference  4,  and  the 
number  of  terms  15.     What  is  the  last  term  ?  Ans.  34. 

3.  The  first  term  is  100,  the  number  of  terms  40,  and  the 
common  difference  2.     What  is  the  last  term  ?  Ans.  22. 

4.  The  first  term  is  80,  the  number  of  terms  10,  and  the 
common  difference  4.     What  is  the  last  term  ?  Ans.  44. 

5.  The  first  term  is  600,  the  number  of  terms  100,  and  the 
common  difference  5.     What  is  the  last  term  ?  Ans.  105. 

6.  The  first  term  is  800,  the  number  of  terms  200,  and  the 
common  difference  2.     What  is  the  last  term  ?  Ans.  402. 

181.  A  progression  by  differences  being  given,  it  is  pro- 
posed to  prove  that  the  su7n  of  any  two  terms,  taken  at  equal 
distances  from  the  tivo  extremes,  is  equal  to  the  sum  of  the 
two  extremes ;  that  is,  if  we  have  the  progression 

2  .  4  .  6  .  8  .  10  .  12, 

we  wish  to  prove  generally  that 

4  +  10,  or  6  +  8, 

is  equal  to  the  sum  of  the  two  extremes,  2  and  12. 

Let  a  .b  .  c  .  e  .f i  .Ic  .1\)q  the  proposed  progression,  and 

n  the  number  of  terms. 

We  will  first  observe  that  if  x  denotes  a  term  which  has 


PROPORTIONS    AND    PROGRESSIONS.  263 

p  terms  before  it,  and  y  a  term  whicli  has  p  terms  after  it, 
we  have,  from  what  has  been  said, 

x=^  a-\-  p  X  d, 

and  3/=  l—p  X  d; 

whence,  by  addition,      x  +  y  =  a  +  l, 

which  proves  the  proposition. 

Referring  to  the  previous  example,  if  we  suppose,  in  the 
first  place,  x  to  denote  the  second  term  4,  then  y  will  denote 
the  term  10,  next  to  the  last.  If  x  denotes  the  third  term  6, 
then  y  will  denote  8,  the  third  term  from  the  last. 

To  apply  this  principle  in  finding  the  sum  of  the  terms 
of  a  progression,  write  the  terms  as  below,  and  then  again 
in  an  inverse  order;  y\z., — 

a  .h  .  c  .  d .  e  ./ i .  Ic  .1. 

I  .  k  ,i c  .b  .  a. 

Calling  8  the  sum  of  the  terms  of  the  first  progression,  28 
will  be  the  sum  of  the  terms  of  both  progressions,  and  we 
shall  have 

28={a+l)  +  (h  +  Jc)  +  {c  +  i).....  +  {i  +  c)  +  (h  +  b)  +  {l+a). 

Now,  since  all  the  parts,  a-\-l,  h-\-h,  c-\-i ,  are  equal 

to  each  other,  and  their  number  equal  to  n, 

28={a  +  r)Xn,  or  8^('^\xn. 

Hence,  for  finding  the  sum  of  an  arithmetical  series,  we 
have  the  following  rule  :  — 

Add  the  tiuo  extremes  together,  and  take  half  their  siim. 
Multiply   this   half  sum   by   the  number   of  terms.      The 
product  will  be  the  sum  of  the  series. 


2G4  ELEMENTARY    ALGEBRA. 


Exercises. 

1.  The  extremes  are  2  and  16,  and  the  number  of  terms  8. 
What  is  the  sum  of  the  series  ? 

S=./^\xn  gives    S=^?^xS  =  72. 

2.  The  extremes  are  3  and  27,  and  the  number  of  terms 
12.     What  is  the  sum  of  the  series?  Ans.  180. 

3.  The  extremes  are  4  and  20,  and  the  number  of  terms 
10.     What  is  the  sum  of  the  series  ?  Ans.  120. 

4.  The  extremes  are  100  and  200,  and  the  number  of  terms 
80.     What  is  the  sum  of  the  series?  Aris.  12000. 

5.  The  extremes  are  500  and  60,  and  the  number  of  terms 
20.     What  is  the  sum  of  the  series  ?  Ajis.  5600. 

6.  The  extremes  are  800  and   1200,  and  the  number  of 
terms  50.     What  is  the  sum  of  the  series  ?  Ans.  50000. 


182.  In  arithmetical  proportion  there  are  five  quantities 
to  be  considered,  —  first,  the  first  term,  a;  second,  the  com- 
mon diflference,  d;  third,  the  number  of  terms,  n;  fourth,  the 
last  term,  I;  fifth,  the  sum,  S. 

The  formulas 

l=a  +  (n-l)d,  and  S  =  f^^^\xn, 

contain  five  quantities,  a,  d,  n,  I,  and  S,  and  consequently 
give  rise  to  the  following  general  problem;  viz., — 

An7/  three  of  these  five  quantities  being  given,  to  determine 
the  other  two. 

We  already  know  the  value  of  S  in  terms  of  a,  n,  and  I. 


PROPORTIONS    AND    PROGRESSIONS.  265 

From  the  formula 

I  —  a  +  (?i  —  1)  c? 
we  find  a  — I  —{n  —  \)d;  that  is, 

y  The  first  term  of  an  increasing  arithmetical  progression  is 
equal  to  the  last  term,  minus  the  product  of  the  common  differ- 
ence by  the  number  of  terms  less  one. 

From  the  same  formula  we  also  find 

d  = ;  that  is, 

n—  1 

In  any  arithmetical  progression,  the  common  difference  is 
equal  to  the  last  term,  minus  the  first  term,  divided  by  the 
number  of  terms  less  one. 

(1)  The  last  term  is  16,  the  first  term  4,  and  the  number 
of  terms  5.     What  is  the  common  difference  ? 

The  formula  d  =  - 


n-l 


-.     16-4      o 
gives  a  =  — - —  =  6. 

(2)  The  last  term  is  22,  the  first  term  4,  and  the  number 
of  terms  10.     What  is  the  common  difference  ?  Ans.  2. 

183.  The  last  principle  affords  a  solution  to  the  following 
question  :  — 

To  find  a  number  m  of  arithmetical  means  between  two 
given  numbers  a  and  b. 

To  resolve  this  question,  it  is  first  necessary  to  find  the 
common  difference.  Now,  we  may  regard  a  as  the  first  term 
of  an  arithmetical  progression,  b  as  the  last  term,  and  the 
required  means  as  intermediate  terms.  The  number  of  terms 
of  this  progression  will  be  expressed  by  m  +  2. 


266  ELEMENTARY    ALGEBRA. 

Now,   by  substituting  in  the  above  formula,  h  for  I,  and 
m  +  2  for  w,  it  becomes 

J  h  —  a  h  —  a      ,1.. 

■ ;  that  IS, 


m  +  2— i      m+1 

TAe  common  difference  of  the  required  progression  is  obtained 
hy  dividing  the  difference  between  the  given  numbers,  a  and  b, 
by  the  required  number  of  means  plus  one. 

Having  obtained  the  common  difference,  d,  form  the  second 
term  of  the  progression,  or  the  first  arithmetical  mean,  by 
adding  d  to  the  first  term  a.  The  second  mean  is  obtained 
by  augmenting  the  first  mean  by  d,  etc. 

(1)  Find  three  arithmetical  means  between  the  extremes 
2  and  18. 

The  formula  rf  =  A_IL^ 

m  +  1 

7     18-2      . 
gives  d  ^ =  4. 

Hence  the  progression  is  2  .  6  .  10  .  14  .  18. 

(2)  Find  twelve  arithmetical  means  between  12  and  77. 

The  formula  d  =  — ^ — 

m  +  1 

gives  d  =  ZlZll?  =  5. 

^  13 

Hence  the  progression  is  12 .  17  .  22  .  27 77. 

184.  If  the  saine  number  of  arithmetical  means  are  inserted 
between  all  the  terms,  taken  two  and  two,  these  terms,  and 
the  arithmetical  means  united,  will  form  one  and  the  same 
progression. 

Let  a  .b  .c  .e ./ be  the  proposed  progression,  and  m  the 

number  of  means  to  be  inserted  between  a  and  b,  b  and  c, 
c  and  e ,  etc. 


PROPORTIONS    AND    PROGRESSIONS.  207 

From  what  has  just  been  said,  the  common  difference  of 
each  partial  progression  will  be  expressed  by 

h  —  a       c  —  h       e  —  c 

>    >    >   } 

m+1     m+1     m  +  1 

expressions  which  are  equal  to  each  other,  since  a,  h,  c,  

are  in  progression.  Therefore  the  common  difference  is  the 
same  in  each  of  the  partial  progressions ;  and,  since  the  last 
term  of  the  first  forms  the  first  term  of  the  second,  etc.,  we 
may  conclude  that  all  of  these  partial  progressions  form  a 
single  progression. 

Exercises. 

1.  Find  the  sum  of  the  first  fifty  terms  of  the  progression 
2.9.16.23 

For  the  50th  term  we  have 

Z  =  2  +  49  X  7  =  345. 
Hence  S={2-\-  345)  x  ^  =  347  X  25  =  8675. 

2.  Find  the  100th  term  of  the  series  2  .  9  .  16  .  23 

Ans.  695. 

3.  Find  the  sum  of  100  terms  of  the  series  1.3.5.7.9 

Ans.  10000. 

4.  The  greatest  term  is  70,  the  common  difference  3,  and 
the  number  of  terms  21.  What  is  the  least  term,  and  the 
sum  of  the  series  ?      Ans.  Least  term,  10  ;  sum  of  series,  840. 

6.  The  first  term  is  4,  the  common  difference  8,  and  the 
number  of  terms  8.  ^.What  is  the  last  termj  and  the  sum  of 
the  series?  Ans.  Last  term,  60;  sum  of  series,  256. 

6.  The  first  term  is  2,  the  last  term  20,  and  the  number 
of  terms  10.     What  is  the  common  difference?  Ans.  2. 

7.  Insert  four  means  between  the  two  numbers  4  and  19. 
What  is  the  series?  Ans.  4  .  7  .  10  .  13  .  16  ..19. 


2G8  ELEMENTARY    ALGEBRA. 

8.  The  first  term  of  a  decreasing  arithinetical  progression 
is  10,  the  common  difference  -J-,  and  the  number  of  terms  21. 
Required  the  sum  of  the  series.  Ans.  140. 

9.  In  a  progression  by  differences,  having  given  the  com- 
mon difference  6,  the  last  term  185,  and  the  sum  of  the  terms 
2945,  find  the  first  term  and  the  number  of  terms. 

Ans.  1st  term,  5  ;  number  of  terms,  31. 

10.  Find  nine  arithmetical  means  between  each  antecedent 
and  consequent  of  the  progression  2  .  5  .  8  .  11  .  14 

Ans.  Common  difference,  or  d,  0.3. 

11.  Find  the  number  of  men  contained  in  a  triangular 
battalion,  the  first  rank  containing  1  man,  the  second  2  men, 
the  third  3,  and  so  on  to  the  nth,  which  contains  n:  in  other 
words,  find  the  expression  for  the  sum  of  the  natural  num- 
bers 1,  2,  3 ,  from  1  to  92  inclusively.         .         o__n(n  +  l) 

2 

12.  Find  the  sum  of  the  n  first  terms  of  the  progression 
of  uneven  numbers  1.3.5.7.9  Ans.  S=  n^. 

13.  One  hundred  stones  being  placed  on  the  ground  in  a 
straight  line  at  the  distance  of  2  yards  apart,  how  far  will 
a  person  travel  who  shall  bring  them  one  by  one  to  a  basket 
placed  at  a  distance  of  2  yards  from  the  first  stone  ? 

Ans.  11  miles,  840  yards. 

Geometrical  Proportion  and  Progression. 

185.  Eatio  is  the  quotient  arising  from  dividing  one  quantity 
by  another  quantity  of  the  same  kind,  regarded  as  a  standard. 
Thus,  if  the  numbers  3  and  6  have  the  same  unit,  the  ratio 
of  3  to  6  will  be  expressed  by 


PROPORTIONS    AND    PROGRESSIONS.  '  269 

and  in  general,  if  A  and  B  represent  quantities  of  tlie  same 
kind,  the  ratio  of  J.  to  -S  will  be  expressed  by 

B 
A 

186.   The  character  oc  indicates  that  one  quantity  varies  as 

another.     Thus, 

Aa:B 

is  read,  ''A  varies  as  ^; "  that  is  to  say,  the  ratio  of  A  to  B, 

T> 

or  — ,  is  constant,  though  A  and  B  themselves  may  change 

value. 

If  there  are  four  numbers,  2,  4,  8,  16,  having  such  values 
that  the  second  divided  by  the  first  is  equal  to  the  fourth 
divided  by  the  third,  the  numbers  are  said  to  form  a  pro- 
portion ;  and  in  general,  if  there  are  four  quantities,  -4,  Bj  (7, 
and  I),  having  such  values  that 

B_I) 

then  A  is  said  to  have  the  same  ratio  to  B  that  C  has  to  D, 
or  the  ratio  of  ^  to  ^  is  equal  to  the  ratio  of  C  to  B.  When 
four  quantities  have  this  relation  to  each  other,  compared 
together  two  and  two,  they  are  said  to  form  a  geometrical 
proportion. 

To  express  that  the  ratio  of  J.  to  ^  is  equal  to  the  ratio 
of  C  to  JD,  we  write  the  quantities  thus  :  — 

A:B::C:B, 
and  read,  "  ^  is  to  ^  as  (7  to  D." 

The  quantities  which  are  compared  the  one  with  the  other 
are  called  terms  of  the  proportion.  The  first  and  the  last 
terms  are  called  the  two  extremes  ;  and  the  second  and  the 
third  terms,  the  two  means.  Thus,  A  and  I)  are  the  extremes ; 
and  J5  and  C  the  means. 


270  ELEMENTARY    ALGEBRA. 

187.  Of  four  terms  of  a  proportion,  the  first  and  the  third 
are  called  the  antecedents ;  and  the  second  and  the  fourth,  the 
consequents ;  and  the  last  is  said  to  be  a  fourth  proportional 
to  the  other  three  taken  in  order.  Thus,  in  the  last  propor- 
tion, A  and  C  are  the  antecedents,  and  B  and  D  the  conse- 
quents. 

188.  Three  quantities  are  in  proportion  when  the  first  has 
the  same  ratio  to  the  second  that  the  second  has  to  the  third ; 
and  then  the  middle  term  is  said  to  be  a  mean  proportional 
between*the  other  quantities.     For  example, 

3:6::6:12; 

and  6  is  a  mean  proportional  between  3  and  12. 

189.  Four  quantities  are  said  to  be  in  proportion  by  inver- 
sion, or  inversely,  when  the  consequents  are  made  the  antece- 
dents, and  the  antecedents  the  consequents. 

Thus,  if  we  have  the  proportion 

3  :  6  :  :  8  :  16, 
the  inverse  proportion  would  be 

6  :  3  :  :  16  :  8. 

190.  Quantities  are  said  to  be  in  proportion  by  alternation, 
or  alternately,  when  antecedent  is  compared  with  antecedent, 
and  consequent  with  consequent. 

Thus,  if  we  have  the  proportion 

3  :  6  :  :  8  :  16, 
the  alternate  proportion  would  be 

3  :  8  :  :  6  :  16. 


PROPORTIONS    AND    PROGRESSIONS.  271 

191.  Quantities  are  said  to  be  in  proportion  by  composition 
when  the  sum  of  the  antecedent  and  consequent  is  compared 
with  either  antecedent  or  consequent. 

Thus,  if  we  have  the  proportion 

2  :*4  :  :  8  :  16, 
the  proportion  by  composition  w^ould  be 

2  +  4  :  2  :  :  8  +  16  :  8  ; 
and  2  +  4:4:  :8  +  16:  16. 

192.  Quantities  are  said  to  be  in  proportion  by  division  when 
the  difference  of  the  antecedent  and  consequent  is  compared 
with  either  antecedent  or  consequent. 

Thus,  if  we  have  the  proportion 

3  :  9  :  :  12  :  36, 
the  proportion  by  division  will  be 

9- 3:3::  36- 12:  12 
and  9-3:9::'36-12:36. 

193.  Equi-mnltiples  of  two  or  more  quantities  are  the  prod- 
ucts which  arise  from  multiplying  the  quantities  by  the  same 
number. 

Thus,  if  we  have  any  two  numbers,  as  6  and  5,  and  multi- 
ply each  of  them  by  any  number,  as  9,  the  equi-multiples  will 
be  54  and  45  :  for 

6  X  9  ==  54,  and  5  X  9  -  45. 

Also  mx  A  and  m  X  B  are  equi-multiples  of  A  and  B, 
the  common  multiplier  being  m. 

194.  Two  quantities,  A  and  B,  which  may  change  their 
values,  are  reciprocally  or  inversely  jprojportional  when  one  is 


272  ELEMENTARY    ALGEBRA. 

proportional  to  unity  divided  hy  the   other,  and   then   their 
product  remains  constant. 

We  express  this  reciprocal  or  inverse  relation  thus : 

in  which  A  is  said  to  be  inversely  proportional  to  B,  or  A 
varies  as  the  reciprocal  of  B, 

195.  If  we  have  the  proportion 

A:B'.:C:D, 

we  have  1  "^  77  (§  ^^^)  ' 

and  by  clearing  the  equation  of  fractions  we  have 

BC=AD;  that  is, 

Of  four  proportional  quantities,  the  product  of  the  two 
extremes  is  equal  to  the  product  of  the  two  means. 

This  general  principle  is  apparent  in  the  proportion  be- 
tween the  numbers 

2  :  10  :  :  12  :  60, 

which  gives  2  X  60  -  10  X  12  -  120. 

196.  If  four  quantities.  A,  B,  C,  D,  are  so  related  to  each 
other  that 

Ax  D  =  BxC, 

we  shall  also  have  —  —  —  ; 

A     C 

and  hence  A:  B  w  C:  D ;  that  is, 

If  the  product  of  two  quantities  is  equal  to  the  product  of 
two  other  quantiti^.s,  two  of  them  may  he  made  the  extremes^ 
and  the  other  two  the  means,  of  a  proportion. 


I 


PROPORTIOXS    AND    PROGRESSIONS.  273 

Thus,  if  we  have        2  X  8  -=  4  X  4, 
we  also  have  2  :  4  :  :  4  :  8. 

197.  If  we  have  three  proportional  quantities, 

.  B      C 

we  nave  —  —  — • 

A      B 

Hence  B^  =  AC;  that  is, 

If  three  quantities  are  proportional,  the  square  of  the  middle 
term  is  equal  to  the  product  of  the  two  extremes. 

Thus,  if  we  have  the  proportion 

3  :  6  :  :  6  :  12, 
we  shall  also  have 

6  X  6  -  6^  =--  3  X  12  =  36. 

198.  If  we  have         A:B::C:D, 

and  consequently  _  r=  _, 

C 
multiply  both  members  of  the  last  equation  by  — ,  and  we 

then  obtain 

and  hence  A:  Cw  B\  D ;  that  is. 

If  four  quantities  arc  proportional,  they  will  he  in  propor- 
tion hy  alternation. 

Let  us  take  as  an  example 

10  :  15  :  :  20  :  30. 
We  shall  have,  by  alternating  the  terms, 

10  :  20  :  :  15  :  30. 

I>.  N.  E.  A.  —  IS. 


274  ELEMENTARY    ALGEBPwA. 

199.  If  we  have 

A:B:.C'.D,  and  A  :  B  :  :  E :  F, 

we  shall  also  have 

B      D        A   ^      ^ 

—  —  —  ,   and   —  —  — -  • 

AC  A      E 

Hence  -  =-  — ,  and  C  :  D  :  :  E :  F ;  that  is, 

C      E 

If  there  are  two  sets  of  pi^opor lions  having  an  antecedent  and 
consequent  of  the  one  equal  to  an  antecedent  and  consequent  of 
the  other,  the  remaining  terms  will  be  proportional. 

If  we  have  the  two  proportions 

2  :  6  :  :  8  :  24,  and  2  :  G  :  :  10  :  30, 

we  shall  also  have 

8  :  24  :  :  10  :  30. 

200.  If  we  have 

7?       7) 

A  \  B  \\  C.  D,  and  consequently  —  =  — » 

AC 

we  have,  by  dividing  1  by  each  member  of  the  equation, 

AC 

-  —  ~,  and  consequently  B  :  A  :  :  B  :  C;  that  is, 

Eour  proportional  quayitities  will  be  in  proportion  when 
taken  inversely. 

To  give  an  example  in  numbers,  take  the  proportion 

7  :  14  :  :  8  :  16. 
Then  the  inverse  proportion  will  be 

14:  7::  16:8, 
in  which  the  ratio  is  one  half. 


PROPORTIONS    AND    PROGRESSIONS.  275 

201.    The  proportion 

A'.BwC'.D  gives  A  X  I)  ^  B  X  C. 

To  each  member  of  the  last  equation  add    B  X  D.     We 
shall  then  have 

(A  +  B)xB=(C+I))xB; 

and  by  separating  the  factors  we  obtain 

A  +  B:B::C+I):D. 

If,  instead  of  adding,  we  subtract  B  X  D  from  both  mem- 
bers, we  have 

{A-B)xD  =  {C--  D)xB, 

which  gives       A  -  B  :  B  :  \  C  -  D  \  D ;  that  is, 

If  four  quantities  are  proportional,  they  will  he  in  proportion 
hy  composition  or  division. 

Thus,  if  we  have  the  proportion 

9  :  27  :  :  16  :  48, 
we  shall  have,  by  composition, 

9  +  27:27:  :  16 +  48:48  • 
that  is,  36  :  27  :  :  64  :  48, 

in  which  the  ratio  is  three  fourths. 

The  same  proportion  gives  us,  by  division-, 
27- 9:  27::  48 -16:  48; 
that  is,  18  :  27  :  :  32  :  48, 

in  which  the  ratio  is  one  and  one  half. 


276  ELEMENTARY    ALGEBEA. 

202.  If  we  have 

J—c' 

and  multiply  the  numerator   and   denominator   of  the   first 
member  by  any  number,  ??i,  we  obtain 

mB  _D 

mA      d 

and  mA  :  mB  \  :  C .  D ;   that  is, 

Uqui-multiples  of  two  quantities  have  the  same  ratio  as  the 
quantities  them^selves. 

For  example  :  if  we  have  the  proportion 

5  :  10  :  :  12  :  24, 

and  multiply  the  first  antecedent  and  consequent  by  6,  we 
have 

30  :  60  :  :  12  :  24, 

in  which  the  ratio  is  still  2. 

203.  The  proportions 

A:B.'.C:D 

and  A:B.:E:F 

give  Ax  D  =  Bx  C, 

and  Ax  F^=Bx  E. 

Adding  and  subtracting  these  equations,  we  obtain 

A(D±F)  =  B(C  ±E), 

or  A:B::C±E:I)±F;  that  is, 

If  C  and  D,  the  antecedent  and  consequent,  be  augmented 
or  diminished  by  quantities  E  and  F,  which  have  the  same 
ratio  as  C  to  D,  the  resulting  quantities  loill  also  have  the 
same  ratio. 


rROrORTIONS    AND    PROGRESSIONS.  277 

Let  us  take  ,as  an  example  the  proportion  9  :  18  :  :  20  :  40, 
in  v/hich  the  ratio  is  2. 

If  we  augment  the  antecedent  and  consequent  by  the 
numbers  15  and  30,  which  have  the  same  ratio,  we  shall  have 
9  +  15  :  18  +  30  :  :  20  :  40 ;  that  is,  24  :  48  :  :  20  :  40,  in  which 
the  ratio  is  still  2. 

If  we  diminish  the  second  antecedent  and  consequent  by 
these  numbers  respectively,  we  have  9  :  18  ::  20  —  15  :  40  —  30  ; 
that  is,  9  :  18  :  :  5  :  10,  in  which  the  ratio  is  still  2. 

204.    If  we  have  several  proportions, 

A'.BwC'.D,  which  gives  A  X  D  =  B  X  C, 
A.BwE.F,   which  gives  A  X  F  =  B  x  U, 
A:  B  ::  G  :  IT,  which  gives  A  X  11=  B  X  G,  etc., 
we  shall  have,  by  addition, 

A(I)  +  F+ir)=^B(C+F+G); 
and  by  separating  the  factors, 

A:B::C+  E+  G:D  +  F+H;  that  is. 

In  any  number  of  proportions  having  the  same  ratio,  any 
antecedent  will  he  to  its  consequent  as  the  sum  of  the  antecedents 
to  the  sum  of  the  consequents. 

Let  us  take,  for  example, 

2  :  4  :  :  6  :  12,  and  1  :  2 :  :  3  :  6,  etc. 

Then  2:4::6  +  3:12  +  6; 

that  is,  2  :  4  :  :  9  :  18, 

in  which  the  ratio  is  still  2. 


278  ELEMENTARY    ALGEBRA. 

205.  If  we  have  four  proportional  quantities, 

A:B::C.D,  we  have  ^  =  ^; 
A      C 

and  raising  both  members  to  any  power  whose  exponent  is  n, 
or  extracting  any  root  whose  index  is  n,  we  have 

1  i_ 

and  consequently 

^"  :  ^"  :  :  C"  :  D\    or    JJ"  \  B^  :  :  C"  :  D^ ;  that  is, 

If  four  quantities  are  proportional,  their  like  powers  or  roots 
will  be  proportional. 

If  we  have,  for  example, 

2  :  4  :  :  3  :  G, 

we  shall  have  2^ :  4^ :  :  3^  6^ 

that  is,  4  :  16  :  :  9  :  36, 

in  which  the  terms  are  proportional,  the  ratio  being  4. 

206.  Let  there  be  two  sets  of  proportions, — 

A  :  B  \\  C  \  D,  which  ejives  —  —  —  ; 
^        A      C 

E\  F wG  '.H,  which  gives  -7=,—  7^* 

Multiplying  them  together  member  by  member,  we  have, 

By.F  _Dy.H 
AxF      CxG' 

Ax  E:  Bx  F::CX  G.BxII;  that  is. 

In  two  sets  of  proportional  quantities,  the  products  of  the 
corresponding  terms  are  proportional. 


PROPORTIONS    AND    PROGRESSIONS.  279 

Thus,  if  we  have  the  two  proportions 

20, 


8:16: 

:10 

and 

3:    4: 

:    6 

we  shall  have 

24 :  64  : 

:60 

160. 

207.  We  have  thus  far  considered  only  the  case  in  which 
the  ratio  of  the  first  term  to  the  second  is  the  same  as  that  of 
the  third  to  the  fourth. 

If  we  have  the  further  condition  that  the  ratio  of  the  second 
term  to  the  third  shall  also  be  the  same  as  that  of  the  first  to 
the  second  or  of  the  third  to  the  fourth,  we  shall  have  a  series 
of  numbers  each  one  of  which,  divided  by  the  preceding  one, 
will  give  the  same  ratio.  Hence,  if  any  term  be  multiplied 
by  this  quotient,  the  product  will  be  the  succeeding  term.  A 
series  of  numbers  so  formed  is  called  a  geometrical  progression. 
Hence 

A  geometrical  progression,  or  progression  by  quotients,  is  a 
series  of  terms  each  of  which  is  equal  to  the  preceding  term 
multiplied  by  a  constant  number,  which  number  is  called  the 
ratio  of  the  progression.     Thus, 

1  :  3  :  9  :  27  :  81  :  243,  etc., 

is  a  geometrical  progression  in  which  the  ratio  is  3.     It  is 
written  by  merely  placing  two  dots  between  the  terms. 

Also  64  :  32  :  16  :  8  :  4  :  2  :  1 

is  a  geometrical  progression  in  which  the  ratio  is  one  half. 

In  the  first  progression  each  term  is  contained  three  times 
in  the  one  that  follows,  and  hence  the  ratio  is  3.  In  the  sec- 
ond, each  term  is  contained  one  half  times  in  the  one  which 
follows,  and  hence  the  ratio  is  one  half. 

The  first  is  called  an  increasing  progression  ;  and  the  second, 
a  decreasing  progression. 


280  ELEMENTARY    ALGEBRA. 

Let  a,  h,  c,  d,  e,f be  numbers  in  a  progression  by  quo- 
tients.    They  are  written  thus  :  — 

a:h  \  c  :  d\  e:f:  g , 

and  it  is  enunciated  in  the  same  manner  as  a  progression  by 
differences.  It  is  necessary,  however,  to  make  the  distinction 
•  that  one  is  a  series  formed  by  equal  differences,  and  the  other 
a  series  formed  by  equal  quotients  or  ratios.  It  should  be 
remarked  that  each  term  is  at  the  same  time  an  antecedent 
and  a  consequent,  except  the  first,  which  is  only  an  antece- 
dent, and  the  last,  which  is  only  a  consequent. 

208.   Let  r  denote  the  ratio  of  the  progression 

a\b  \  c  \  d , 

r  being  >  1  when  the  progression  is  increasing,  and   r  <  1 
when  it  is  decreasing.     Then,  since 


we  have 


b  c  d  e  , 

_  —  r,  -  —  r,  -  ^  r,  -  =  r,    etc., 
abed 

b  —  ar,    c  =  br  —  ar"^,    d~cr  =  ar^, 

e  =  dr  =  ar^,  f=  er  ^=  ar^, ; 

that  is,  the  second  term  is  equal  to  ar,  the  third  to  ar"^,  the 
fourth  to  ar^j  the  fifth  to  ar*,  etc. ;  and  in  general  the  nth 
term,  that  is,  one  which  has  7i—  1  terms  before  it,  is  expressed 
by  ar"*"'. 

Let  I  be  this  term.     "We  then  have  the  formula 

lz=zar''-\ 

by  means  of  which  we  can  obtain  any  term  without  being 
obliged  to  find  all  the  terms  which  precede  it.  Hence,  to 
find  the  last  term  of  a  progression,  we  have  the  following 
rule. 


PROPORTIONS    AND    PROGRESSIONS.  281 

Raise  the  ratio  to  a  power  whose  exponent  is  one  less  than 
the  number  of  terms. 

Multiply  the  power  thus  found  by  the  first  term.  The  prod- 
uct will  be  the  required  term. 

Exercises. 

1.  Find  the  5th  term  of  the  progression  2  :  4  :  8  :  16  ....* 
in  which  the  first  term  is  2,  and  the  common  ratio  2. 

5th  term  =  2  X  2*  --=  2  X  16  =  32,  Ans. 

2.  Find  the  8th  term  of  the  progression  2  :  6  :  18  :  54 

8th  term  -  2  X  3^  =-  2  X  2187  -  4374,  Ans. 

3.  Find  the  6th  term  of  the  progression  2  :  8  :  32  :  128 

6th  term  ==  2  X  4^  ==  2  X  1024  =-  2048,  Ans. 

4.  Find  the  7th  term  of  the  progression  3  :  9  :  27  :  81 

7th  term  -=  3  X  3'  ==  3  X  729  =  2187,  Ans. 

5.  Find  the  6th  term  of  the  progression  4  :  12  :  36  :  108 

6th  term  --  4  X  3^  -  4  X  243  -  972,  Ans. 

6.  A  person  agreed  to  pay  his  servant  one  cent  for  the  first 
day,  two  for  the  second,  and  four  for  the  third,  doubling 
every  day  for  ten  days.  How  much  did  he  receive  on  the 
tenth  day?  Ans.  $5.12. 

7.  What  is  the  8th  term  of  the  progression 

9:36:144:576 ? 

8th  term  .==  9  X  4^  =  9  X  16384  =  147456,  Ans. 

8.  Find  the  12th  term  of  the  progression 

64:16:4:l:i 

4 

12th  term- 64 /'-V=-£  =  i-—l—,    Ans. 
V4y       4^^      4«     65536 


282  ELEMENTARY    ALGEBRA. 

209.    We  will  now  proceed  to  determine  the  sum  of  n  terms 
of  a  progression, 

a\h  '.c\  d\e:f: \i\k:l, 

I  denoting  the  nth  term. 

We  have  the  equations  (§  208) 

b  =  ar^  c  =  hr^  d  =  cr,  e  =  dr, Jc=  ir,  1=  kr ; 

and  by  adding  them  all  together,  member  to  member,  we 
deduce 

Sum  of  1st  Members.                                Sum  of  2d  Members. 
b-\-c+d+e  + -\.]c+l  =  {a  +  b  +  c  +  d+ +  i  +  k)r, 

in  which  we  see  that  the  first  member  contains  all  the  terms 
but  a,  and  the  polynomial  within  the  parenthesis  in  the  second 
member  contains  all  the  terms  but  I.  Hence,  if  we  call  the 
sum  of  the  terms  8,  we  have 

S-a-^{S-l)r:=8r-lr, 
:.  Sr-8=lr-a; 
It  —  a 


whence  xS'- 


r-1 


Therefore,  to  obtain  the  sum  of  all  the  terms,  or  sum  of  the 
series  of  a  geometrical  progression,  we  have  the  rule  :  — 

Multiply  the  last  term  by  the  ratio. 

Sabtraci  the  first  term  from  the  product. 

Divide  the  remainder  by  the  ratio  diminished  by  1,  and  the 
quotient  will  be  the  sum  of  the  series. 

Exercises. 

1.    Find  the  sum  of  eight  terms  of  the  progression 
2:6:18:54:162 2  X  S' =  4374. 

O  Ir-a  13122-2  n.nr.         . 

;^— ^  —  — -_^ — .^  =  6560,  ^725. 

r-l  2 


PROPORTIONS    AND    PROGRESSIONS.  283 

2.  Find  the  sum  of  the  progression  2  :  4  :  8  :  16  :  32. 

o      It  ~  a      64  —  2      no     a 

S  = -  =  — :; — =^62,  Ans. 

r  —  1  1 

3.  Find  the  sum  of  ten  terms  of  the  progression 

2  :  6  :  18  :  54  :  162 2x3^  =  39366. 

Ans.  59048. 

4.  What  debt  may  be  discharged  in  a  year,  or  twelve 
months,  by  paying  $1  the  first  month,  $2  the  second  month, 
$4  the  third  month,  and  so  on,  each  succeeding  payment 
being  double  the  last ;  and  what  will  be  the  last  payment  ? 

Ans.  Debt,  $4095  ;  last  payment,  $2048. 

5.  A  daughter  was  married  on  New- Year's  Day.  Her 
father  gave  her  Is.,  with  an  agreement  to  double  it  on  the 
first  of  the  next  month,  and  at  the  beginning  of  each  succeed- 
ing month  to  double  what  she  had  received  the  previous  month. 
How  much  had  she  received  at  the  end  of  the  year  ? 

.4ns.  £204  15s. 

6.  A  man  bought  10  bushels  of  wheat  on  the  condition' 
that  he  should  pay  1  cent  for  the  first  bushel,  3  for  the  second, 
9  for  the  third,  and  so  on  to  the  last.  What  did  he  pay  for 
the  last  bushel,  and  what  for  the  ten  bushels? 

Ans.  Last  bushel,  $196.83  ;  total  cost,  $295.24. 

7.  A  man  plants  4  bushels  of  barley,  which  at  the  first 
harvest  produced  32  bushels ;  these  he  also  plants,  which,  in 
like  manner,  produce  eightfold ;  he  again  plants  all  his  crop, 
and  again  gets  eightfold ;  and  so  on  for  16  years.  What  is 
his  last  crop,  and  what  the  sum  of  the  series  ? 

Ans.  Last,  140,737,488,355,328  bushels  ; 
sum,  160,842,843,834,660. 


284  ELEMENTARY    ALGEBRA. 

210.    When  the  progression  is  decreasing,  we  have  r<l, 
and  l<a.     The  above  formula, 

a      Ir  —  a 
r  —  1 

for  the  sum,  is  then  written  under  the  form 

1  —  r 

in  order  that  the  two  terms  of  the  fraction  may  be  positive. 

Exercises. 

1.  Find  the  sum  of  the  terms  of  the  progression 

32  :  16  :  8  :  4  :  2. 

,       32-2x^      ^, 
o     a- It     2     31      „      . 

1  —  r  1  1 

2  2 

2.  Find  the  sum  of  the  first  twelve   terms   of  the   pro- 
gression 

64:16:4:1:^: :64(^-y,  or       ^ 


4  VV  65536 

64---L:-Xt     256--    ^ 


cr„^-^^_  6^^^6     ^_  65^36  _^^  ,    65535 

^"r=7 3  ■ 3  ^^  +  196608' 

4 

211.  Note.  —  We  perceive  that  the  principal  difficulty  consists  in 
obtaining  the  numerical  value  of  the  last  term,  —  a  tedious  operation, 
even  when  the  number  of  terms  is  not  very  great. 

3.    Find  the  sum  of  six  terms  of  the  progression 

512:128:32 

Ans.  682f 


PROPORTIONS   AND    PROGRESSIONS.  285 

4.  Find  the  sum  of  seven  terms  of  the  progression 

2187  :  729  :  243 

Ans.  3279. 

5.  Find  the  sum  of  six  terms  of  the  progression 

972  :  324  :  108 

Ans.  1456. 

6.  Find  the  sum  of  eight  terms  of  the  progression 

147456  :  36864  :  9216 

Ans.  196605. 

212.    Let  there  be  the  decreasing  progression 

a  :  b  :  c  :  d :  e  :f  : , 

containing  an  indefinite  number  of  terms.     In  the  formula 

o     a  —  It 

^^'\ 

1  —  r 

substitute  for  I  its  value  ar"~^  (§  208),  and  we  have 

a  —  ar"^ 


S  =  ' 


1 


which  expresses  the  sum  of  ?i  terms  of  the  progression.     This 
may  be  put  under  the  form 


1  —  r      1  —  r 

Now,  since  the  progression  is  decreasing,  r  is  a  proper  frac- 
tion ;  and  r"*  is  also  a  fraction,  which  diminishes  as  n  increases. 
Therefore  the  greater  the  number  of  terms  we  take,  the  more 
will  - — —  X  r"  diminish,  and  consequently  the  more  will  the 
entire  sum  of  all  the  terms  approximate  to  an  equality  with 

the  first  part  of  S;    that  is,  to  — '■ — .     Finally,  when  n  is 

1  —  r 

taken  greater  than  any  given  number,  or  n  =  infinity,  then 


286  ELEMENTARY    ALGEBRA. 

X  r**  will  be  less  than  any  given  number,  or  will  differ 

so  little  from  0  that  its  value  may  be  disregarded  without 

making  any  appreciable  difference  in  the   result ;    and   the 

expression will  then  represent  the  true  value  of  the  sum 

1  —  r 

of  all  the  terms  of  the  series.  Whence  we  may  conclude  that 
the  expression  for  the  sum  of  the  terms  of  a  decreasing  progres- 
sion^ in  which  the  number  of  terms  is  infinite,  is 


1-r 

that  is,  equal  to  the  first  teryn,  divided  by  1  7ninus  the  ratio. 

That  is,  properly  speaking,  the  limit  to  which  the  partial 

sums  approach,  as  we  take  a  greater  number  of  terms  in  the 

progression.     The   difference   between   these  sums  and 

1  —  r 

may  be  made  as  small  as  we  please,  but  will  only  become 
inappreciable  when  the  number  of  terms  is  infinite. 

Exercises. 

1.    Find  the  sum  of  1  :  -  :  -  :  —  :  —  to  infinity. 
3   9   27   81  ^ 

We  have,  for  the  expression  of  the  sum  of  the  terms, 
>S=-~^  =  -i-.  =  ^=U,  Alls. 

3 
The  error  committed  by  taking  this  expression  for  the  value  of  the 
sum  of  the  n  first  terms  is  expressed  by 


1  — r 


First  take  n  =  5,  it  becomes 

;wiy^    1    ^   1 
2V3/      2.3*      16 


162 


PROPORTIONS    AND    PROGRESSIONS.  287 

When  71  =  6,  we  find 


1\3/       162     3      486 


3 
The  error  committed  by  taking  -  for  the  sum  of  a  number  of  terms 

is  evidently  less  in  proportion  as  the  number  of  terras  is  greater. 

2.  Again  take  the  progression 

^.i.i.i.2_._i_. 

*2'4"8"1G'32*^  ^ 

8  = ==^  — r  =  2,  Ans, 

l~r      ^_1 

2 

3.  What  is  the  sum  of  the  progression 

1,    — >    '    J    ,  etc.,  to  infinity? 

'    10     100     1000     10000         '  ^ 

S  =  z: ^ =  11,  Ans. 

1  —  /•       -J J_        ^ 

10 

213.  In  the  several  questions  of  geometrical  progression 
there  are  five  numbers  to  be  considered,  —  first,  the  first  term, 
a;  second,  the  ratio,  r;  third,  the  number  of  terms,  n; 
fourth,  the  last  term,  I ;  fifth,  the  sum  of  the  terms,  S. 

214.  We  shall  terminate  this  subject  by  solving  this  prob- 
lem :  — 

Find  a  mean  proportional  between  any  two  numbers,  as 
m  and  n. 

Denote  the  required  mean  by  x.    We  shall  then  have  (§  197) 
x"^  —  mX  n, 


and  hence  x  =  Vm  X  n;  that  is. 

Multiply  the  tivo  numbers  together,  and  extract  the  square 
root  of  the  product. 


ELEMENTARY    ALGEBRA. 


Exercises. 


1.  "What  is  the  geometrical   mean    between    the    numbers 
2  and  8? 

Mean  ^  VS  X  2  =^  VT6  =  4,  Ajis. 

2.  What  is  the  mean  between  4  and  16  ?  Ans.  8. 

3.  What  is  the  mean  between  3  and  27  ?  Ans.  9. 

4.  What  is  the  mean  between  2  and  72?  Ans.  12. 

5.  What  is  the  mean  between  4  and  64?  Ans.  16. 


CHAPTER  X. 

LOGARITHMS. 

215.  The  nature  and  properties  of  the  logarithms  in  com- 
mon use  will  be  readily  understood  by  considering  attentively 
the  different  powers  of  the  number  10.     They  are 

10'' =1, 

10^  -  10, 

10"'  -  100, 

10^  =  1000, 

10*  =  10000, 

10^  =  100000,  etc. 

It  is  plain  that  the  exponents  0,  1,  2,  3,  4,  5,  etc.,  form  an 
arithmetical  progression  of  which  the  common  difference  is  1 ; 
and  that  the  corresponding  numbers  1,  10,  100,  1000,  10000, 
100000,  etc.,  form  a  geometrical  progression  of  which  the 
common  ratio  is  10.  The  number  10  is  called  the  base  of  the 
system  of  logarithms;  and  the  exponents  0,  1,  2,  3,  4,  5,  etc., 
are  the  logarithms  of  the  numbers  whicb  are  produced  by 
raising  10  to  the  powers  denoted  by  those  exponents. 

216.  If  we  denote  the  logarithm  of  any  number  by  m,  then 
the  number  itself  will  be  the  mth  power  of  10  ;  that  is,  if  we 
represent  the  corresponding  number  by  M, 

10"*  =  3f. 

mm      Thus,  if  we  make  m  =  0,  J/ will  be  equal  to  1 ;  if  m  =  1, 
M  will  be  equal  to  10  ;  etc.     Hence 

I).  N.  K.  A.~19.  2S9 


290  ELEMENTARY    ALGEBRA. 

The  logarithm  of  a  number  is  the  exponent  of  the  power  to 
lohich  it  is  necessary  to  raise  the  base  of  the  system  in  order  to 
produce  the  number. 

217.  If,  as  before,  10  denotes  the  base  of  the  system  of 
logarithms,  m  any  exponent,  and  M  the  corresponding  num- 
ber, we  shall  then  have 

10"*  =  J/  (1) 

in  which  ni  is  the  logarithm  of  M. 

If  we  take  a  second  exponent,  n,  and  let  N  denote  the  cor- 
responding number,  we  shall  have 

10"  =  iV^  .  (2) 

in  which  n  is  the  logarithm  of  N, 

If,  now,  we  multiply  the  first  of  these  equations  by  the 
second,  member  by  member,  we  have 

10"*  X  10"  =  10"*+"  -  3/x  N. 

But,  since  10  is  the  base  of  the  system,  ni  +  n  is  the  loga- 
rithm of  Mx  N.     Hence 

The  sum  of  the  logarithms  of  any  two  numbers  is  equal  to 
the  logarithm  of  their  product. 

Therefore  the  addition  of  logarithms  corresponds  to  the  mul- 
tiplication of  their  numbers. 

218.  If  we  divide  Equation  (1)  by  Equation  (2),  member 
by  member,  we  have 

lo:*  N 


But,  since  10  is  the  base  of  the  system,  ni  —  n  is  the  loga- 
M 

N 


M 

rithm  of  — ••     Hence 


LOGARITHMS.  291 

If  one  number  he  divided  by  another,  the  logarUhni  of  the 
quotient  will  be  equal  to  the  logarithm  of  the  dividend,  dimin- 
ished by  that  of  the  divisor. 

Therefore  the  subtraction  of  logarithms  corresponds  to  the 
division  of  their  numbers. 

219.    Let  us  examine  further  the  equations 

10«-1, 
10^-10, 
10^  -  100, 
10-^  -  1000,  etc. 

It  is  plain  that  the  logarithm  of  1  is  0,  and  that  the  loga- 
rithm of  any  number  between  1  and  10  is  greater  than  0  and 
less  than  1.  The  logarithm  is  generally  expressed  by  decimal 
fractions.     Thus, 

log  2 -0.301030. 

The  logarithm  of  any  number  greater  than  10  and  less 
than  100  is  greater  than  1  and  less  than  2,  and  is  expressed 
by  1  and  a  decimal  fraction.     Thus, 

log  50  =1.698970. 

The  part  of  the  logarithm  which  stands  at  the  left  of  the 
decimal  point  is  called  the  characteristic  of  the  logarithm,  and 
the  part  that  stands  at  the  right  of  the  decimal  point  is  called 
the  mantissa  of  the  logarithm.  In  a  whole  number,  the  char- 
acteristic is  always  one  less  than  the  mimber  of  places  of  figures 
in  the  number  whose  logarithm  is  taken. 

Thus,  in  the  first  case,  for  numbers  bj^tween  1  and  10  there 
is  but  one  place  of  figures,  and  the  characteristic  is  0;  for 
numbers  between  10  and  100  there  are  two  places  of  figures, 
and  the  characteristic  is  1 ;  and  similarly  for  other  numbers. 

If  we   form  the  powers  of  10  denoted  by  negative  expo- 


292  ELEMENTARY    ALGEBRA. 

nents  (see  §  50),  we  have,  from  the  definition  of  a  logarithm, 
the  following  table  :  — 


10 

1 

100 

1 
1000 


(ior=T^  =-01;    •••  i°g-oi  =-2- 


aor=T^=-001;     .-.log  .001  =  -3. 


And  so  on. 


Here  we  see  that  the  logarithm  of  every  number  between 
1  and  .1  is  found  between  0  and  —  1 ;  that  is,  it  is  equal  to 

—  1,  plus  a  part  less  than  1.  The  logarithm  of  every  num- 
ber between  .1  and  .01  is  between  —  1  and  —2;  that  is,  it 
is  equal  to  —  2,  plus  a  part.  The  logarithm  of  every  num- 
ber between  .01  and  .001  is  between  —  2  and  —  3,  or  it  is 
equal  to  —  3,  plus  a  part.     And  so  on. 

In  the  first  case  the  characteristic  is  —  1 ;  in  the  second, 

—  2  ;  in  the  third,  —  3  ;  and  so  on.     That  is, 

The  characteristic  of  the  logarithvi  of  a  decimal  is  negative, 
and  numerically  one  greater  than  the  number  of  ciphers  that 
immediately  follow  the  decimal  point. 

In  a  mixed  number,  the  characteristic  of  its  logarithm  will 
evidently  be  one  less  than  the  number  of  places  of  figures  in 
its  entire  part. 

The  decimal  part  of  a  logarithm  is  always  regarded  as 
positive ;  and,  to  indicate  that  the  negative  sign  extends  only 
to  the  characteristic*  it  is  generally  written  over  it.     Thus, 

log  0.012  =  2.079181,  equivalent  to  -2  prefixed  to  +.079181. 

220.  A  table  of  logarithms  is  a  table  containing  a  set  of 
numbers  and  their  logarithms,  so  arranged  that  we  may,  by 


LOGARITHMS. 


293 


its  aid,  find  the  logarithm  corresponding  to  any  number  of 
the  set  or  the  number  corresponding  to  any  of  the  logarithms. 
A  table  showing  the  logarithms  of  all  whole  numbers  from 
1  to  100  is  annexed.  The  numbers  are  written  in  the  column 
designated  by  the  letter  '' N,"  and  the  logarithms  in  the 
column  designated  by  "  Log." 

Table. 


N. 

Log. 

N. 

Log. 

N. 

Log. 

N. 

Log. 

1 

0.000000 

26 

1.414973 

51 

1.707570 

76 

1.880814 

2 

0.301030 

27 

1.431364 

52 

1.716003 

77 

1.886491 

3 

0.477121 

28 

1.447158 

53 

1.724276 

78 

1.892095 

4 

0.602060 

29 

1.462398 

54 

1.732394 

79 

1.897627 

5 

0.698970 

30 

1.477121 

55 

1.740363 

80 

1.903090 

6 

0.778151 

31 

1.491362 

56 

1.748188 

81 

1.908485 

7 

0.845098 

32 

1.505150 

57 

1.755875 

82 

1.913814 

8 

0.903090 

33 

1.518514 

58 

1.763428 

83 

1.919078 

9 

0.954243 

34 

1.531479 

59 

1.770852 

84 

1.924279 

10 

1.000000 

35 

1.544068 

60 

1.778151 

85 

1.929419 

11 

1.041393 

36 

1.556303 

61 

1.785330 

86 

1.934498 

12 

1.079181 

37 

1.568202 

62 

1.792392 

87 

1.939519 

13 

1.113943 

38 

1.579784 

63 

1.799341 

88 

1.944483 

14 

1.146128 

39 

1.591065 

64 

1.806180 

89 

1.949390 

15 

1.176091 

40 

1.602060 

Qb 

1.812913 

90 

1.954243 

16 

1.204120 

41 

1.621784 

66 

1.819544 

91 

1.959041 

17 

1.230449 

42 

1.623249 

67 

1.826075 

92 

1.963788 

18 

1.255273 

43 

1.633468 

68 

1.832509 

93 

1.968483 

19 

1.278754 

44 

1.643453 

69 

1.838849 

94 

1.973128 

20 

1.301030 

45 

1.653213 

70 

1.845098 

95 

1.977724 

21 

1.322219 

46 

1.662758 

71 

1.851258 

96 

1.982271 

22 

1.342423 

47 

1.672098 

72 

1.857333 

97 

1.986772 

23 

1.361728 

48 

1.681241 

73 

1.863323 

98 

1.991226 

24 

1.380211 

49 

1.690196 

74 

1.869232 

99 

1.995635 

25 

1.397940 

50 

1.698970 

75 

1.875061 

100 

2.000000 

Exercises. 


1.    Let   it   be    required   to  multiply  8  by  9  by  means  of 


logarithms. 


294  ELEMENTARY    ALGEBRA. 

Note.  —  We  have  seen  (§  216)  that  the  sum  of  the  logarithms  is  equal 
to  the  logarithm  of  the  product :  therefore  find  the  logarithm  of  8  from 
the  table,  which  is  0.903090  ;  and  then  the  logarithm  of  9,  which  is 
0.954243  ;  and  their  sum,  which  is  1.857333,  will  be  the  logarithm  of 
the  product.  In  searching  along  in  the  table,  we  find  that  72  stands 
opposite  this  logarithm  :  hence  72  is  the  product  of  8  by  9. 

2.  What  is  the  product  of  7  by  12  ? 

Logarithm  of  7 0.845098 

Logarithm  of  12     . 1.079181 

Logarithm  of  their  product 1.924279 

Corresponding  number,  84. 

3.  What  is  the  product  of  9  by  11  ? 

Logarithm  of  9 0.954243 

Logarithm  of  11 1.041393 

Logarithm  of  their  product 1.995636 

Corresponding  number,  99. 

4.  Let  it  be  required  to  divide  84  by  3. 

Note.  —  We  have  seen  (§  218)  that  the  subtraction  of  logarithms 
corresponds  to  the  division  of  their  numbers  :  hence  if  we  find  the 
logarithm  of  84,  and  then  subtract  from  it  the  logarithm  of  3,  the 
remainder  will  be  the  logarithm  of  the  quotient. 

Logarithm  of  84 1.924279 

Logarithm  of  3 •     •     •     0.477121 

Difference 1.447158 

Corresponding  number,  28. 

5.  Divide  42  by  7. 

Logarithm  of  42 1.623249 

Logarithm  of  7 0.845098 

Logarithm  of  the  quotient 0.778151 

Corresponding  number,  6. 

OF  THE 
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